【小白刷leetcode】第15题
动手刷leetcode，正在准备蓝桥，但是本人算法能力一直是硬伤。。。所以做得一直很痛苦。但是不熟练的事情像练吉他一样，就需要慢速，多练。
题目描述
看这个题目，说实在看的不是很懂。索性我们直接来看输入输出。观察输入输出我们可以知道：输入是一个数组，输出是一个list<list。目标其实就是要找到其中的三元组，三元组可以加起来等于零。而且用过的来组成三元组的不能重复使用。也就是每个元素只能使用一次。
暴力解法
首先，我先想到了暴力破解。
每一个元素都访问一遍，用三个for循环。未来防止ijk访问重复，让每一的初始值都为变化的。j=i+1，k=j+1。
这个理论上是行的通的，就是时间超过了限制。时间复杂度为o(n^3)。然后到了一个测试样例过不了。我用idea试了那个测试样例一直在跑，结束不了了。。。
public static List<List<Integer>> threeSum(int[] nums) {
//初步看输入输出理解题意，因为我看题目描述压根看不明白
//输入一个数组，输出三元数组，里面包含的list，每个加起来等于零，而且使用元素不能重复
int mask =0;
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
for (int j = i+1; j < nums.length; j++) {
for (int k = j+1; k < nums.length; k++) {
// mask = nums[i]+nums[j];
if((nums[i]+nums[j]+nums[k])==0){
List<Integer> list = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[k]));
Collections.sort(list);
result.add(list);
}
}
}
}
Set<List<Integer>> set = new HashSet<>(result);
List<List<Integer>> uniqueList = new ArrayList<>(set);
return uniqueList;
}
解释一下代码，就是把符合条件的list经过排序放到list中，经过排序是可以后面用来去重。然后添加完了之后，就把list转化为set集合，然后多余的元素都会被去除，然后由集合再转化为list就可以得到最终结果。
测试的mian函数
下面是测试用的main函数。
public static void main(String[] args) {
long start = System.nanoTime();
// int[] nums = {-1,0,1,2,-1,-4};
// int[] nums = {82597,-9243,62390,83030,-97960,-26521,-61011,83390,-38677,12333,75987,46091,83794,19355,-71037,-6242,-28801,324,1202,-90885,-2989,-95597,-34333,35528,5680,89093,-90606,50360,-29393,-27012,53313,65213,99818,-82405,-41661,-3333,-51952,72135,-1523,26377,74685,96992,92263,15929,5467,-99555,-43348,-41689,-60383,-3990,32165,65265,-72973,-58372,12741,-48568,-46596,72419,-1859,34153,62937,81310,-61823,-96770,-54944,8845,-91184,24208,-29078,31495,65258,14198,85395,70506,-40908,56740,-12228,-40072,32429,93001,68445,-73927,25731,-91859,-24150,10093,-60271,-81683,-18126,51055,48189,-6468,25057,81194,-58628,74042,66158,-14452,-49851,-43667,11092,39189,-17025,-79173,13606,83172,92647,-59741,19343,-26644,-57607,82908,-20655,1637,80060,98994,39331,-31274,-61523,91225,-72953,13211,-75116,-98421,-41571,-69074,99587,39345,42151,-2460,98236,15690,-52507,-95803,-48935,-46492,-45606,-79254,-99851,52533,73486,39948,-7240,71815,-585,-96252,90990,-93815,93340,-71848,58733,-14859,-83082,-75794,-82082,-24871,-15206,91207,-56469,-93618,67131,-8682,75719,87429,-98757,-7535,-24890,-94160,85003,33928,75538,97456,-66424,-60074,-8527,-28697,-22308,2246,-70134,-82319,-10184,87081,-34949,-28645,-47352,-83966,-60418,-15293,-53067,-25921,55172,75064,95859,48049,34311,-86931,-38586,33686,-36714,96922,76713,-22165,-80585,-34503,-44516,39217,-28457,47227,-94036,43457,24626,-87359,26898,-70819,30528,-32397,-69486,84912,-1187,-98986,-32958,4280,-79129,-65604,9344,58964,50584,71128,-55480,24986,15086,-62360,-42977,-49482,-77256,-36895,-74818,20,3063,-49426,28152,-97329,6086,86035,-88743,35241,44249,19927,-10660,89404,24179,-26621,-6511,57745,-28750,96340,-97160,-97822,-49979,52307,79462,94273,-24808,77104,9255,-83057,77655,21361,55956,-9096,48599,-40490,-55107,2689,29608,20497,66834,-34678,23553,-81400,-66630,-96321,-34499,-12957,-20564,25610,-4322,-58462,20801,53700,71527,24669,-54534,57879,-3221,33636,3900,97832,-27688,-98715,5992,24520,-55401,-57613,-69926,57377,-77610,20123,52174,860,60429,-91994,-62403,-6218,-90610,-37263,-15052,62069,-96465,44254,89892,-3406,19121,-41842,-87783,-64125,-56120,73904,-22797,-58118,-4866,5356,75318,46119,21276,-19246,-9241,-97425,57333,-15802,93149,25689,-5532,95716,39209,-87672,-29470,-16324,-15331,27632,-39454,56530,-16000,29853,46475,78242,-46602,83192,-73440,-15816,50964,-36601,89758,38375,-40007,-36675,-94030,67576,46811,-64919,45595,76530,40398,35845,41791,67697,-30439,-82944,63115,33447,-36046,-50122,-34789,43003,-78947,-38763,-89210,32756,-20389,-31358,-90526,-81607,88741,86643,98422,47389,-75189,13091,95993,-15501,94260,-25584,-1483,-67261,-70753,25160,89614,-90620,-48542,83889,-12388,-9642,-37043,-67663,28794,-8801,13621,12241,55379,84290,21692,-95906,-85617,-17341,-63767,80183,-4942,-51478,30997,-13658,8838,17452,-82869,-39897,68449,31964,98158,-49489,62283,-62209,-92792,-59342,55146,-38533,20496,62667,62593,36095,-12470,5453,-50451,74716,-17902,3302,-16760,-71642,-34819,96459,-72860,21638,47342,-69897,-40180,44466,76496,84659,13848,-91600,-90887,-63742,-2156,-84981,-99280,94326,-33854,92029,-50811,98711,-36459,-75555,79110,-88164,-97397,-84217,97457,64387,30513,-53190,-83215,252,2344,-27177,-92945,-89010,82662,-11670,86069,53417,42702,97082,3695,-14530,-46334,17910,77999,28009,-12374,15498,-46941,97088,-35030,95040,92095,-59469,-24761,46491,67357,-66658,37446,-65130,-50416,99197,30925,27308,54122,-44719,12582,-99525,-38446,-69050,-22352,94757,-56062,33684,-40199,-46399,96842,-50881,-22380,-65021,40582,53623,-76034,77018,-97074,-84838,-22953,-74205,79715,-33920,-35794,-91369,73421,-82492,63680,-14915,-33295,37145,76852,-69442,60125,-74166,74308,-1900,-30195,-16267,-60781,-27760,5852,38917,25742,-3765,49097,-63541,98612,-92865,-30248,9612,-8798,53262,95781,-42278,-36529,7252,-27394,-5021,59178,80934,-48480,-75131,-54439,-19145,-48140,98457,-6601,-51616,-89730,78028,32083,-48904,16822,-81153,-8832,48720,-80728,-45133,-86647,-4259,-40453,2590,28613,50523,-4105,-27790,-74579,-17223,63721,33489,-47921,97628,-97691,-14782,-65644,18008,-93651,-71266,80990,-76732,-47104,35368,28632,59818,-86269,-89753,34557,-92230,-5933,-3487,-73557,-13174,-43981,-43630,-55171,30254,-83710,-99583,-13500,71787,5017,-25117,-78586,86941,-3251,-23867,-36315,75973,86272,-45575,77462,-98836,-10859,70168,-32971,-38739,-12761,93410,14014,-30706,-77356,-85965,-62316,63918,-59914,-64088,1591,-10957,38004,15129,-83602,-51791,34381,-89382,-26056,8942,5465,71458,-73805,-87445,-19921,-80784,69150,-34168,28301,-68955,18041,6059,82342,9947,39795,44047,-57313,48569,81936,-2863,-80932,32976,-86454,-84207,33033,32867,9104,-16580,-25727,80157,-70169,53741,86522,84651,68480,84018,61932,7332,-61322,-69663,76370,41206,12326,-34689,17016,82975,-23386,39417,72793,44774,-96259,3213,79952,29265,-61492,-49337,14162,65886,3342,-41622,-62659,-90402,-24751,88511,54739,-21383,-40161,-96610,-24944,-602,-76842,-21856,69964,43994,-15121,-85530,12718,13170,-13547,69222,62417,-75305,-81446,-38786,-52075,-23110,97681,-82800,-53178,11474,35857,94197,-58148,-23689,3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int[] nums 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// System.out.println(threeSum(nums));
System.out.println(threeSum8(nums));
Util.printCostTime(start);
// System.out.println(binarySearch(nums,0, nums.length, 2));
}
第一次优化
然后我就开始尝试优化，肯定要优化掉一层循环,于是我就把最后一层循环用了顺序查找代替，因为一开始数组是无序的，我便想到排序可以优化查询性能，速度是优化了，但是还是过不了很大很多的测试样例。
public static List<List<Integer>> threeSum2(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
for (int j = i+1; j < nums.length; j++) {
int find = SequentialSearch(nums,(0-nums[i]-nums[j]));
if (find!= -1&&find !=i&&find !=j){
List<Integer> list = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[find]));
Collections.sort(list);
result.add(list);
}
}
}
Set<List<Integer>> set = new HashSet<>(result);
List<List<Integer>> uniqueList = new ArrayList<>(set);
return uniqueList;
}
public static int SequentialSearch(int[] nums, int target){
for (int i = 0; i < nums.length; i++) {
if(nums[i]==target){
return i;
}
}
return -1;
}
第二次优化
既然排序了，顺序查询不用排序也可以，而且如果使用二分查找的话更加快得离谱，直接把时间复杂度的一个n转化成logn了。
public static List<List<Integer>> threeSum4(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
for (int j = nums.length-1; j > i+1; j--) {
int find = binarySearch(nums,i+1,j,(0-nums[i]-nums[j]));
if (find!= -1&&find !=i&&find !=j){
List<Integer> list = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[find]));
Collections.sort(list);
result.add(list);
}
}
}
Set<List<Integer>> set = new HashSet<>(result);
List<List<Integer>> uniqueList = new ArrayList<>(set);
return uniqueList;
}
- 效果是过了一些前面过不了的测试样例，但是来到3000个0的时候运行时间超过了4秒钟，无疑又是失败的。
第三次优化
我经过上一次的优化，得到了一个想法：我的for循环块好像已经优化得可以了，3000个0不行，肯定是这种数据太怪了，有没有办法把数据经过清洗然后再传入呢。于是，这两次的优化都是往数据清洗方面靠了。
public static List<List<Integer>> threeSum1(int[] nums) {
Set<Integer> nums1 = new HashSet<>();
for (int num : nums) {
nums1.add(num);
}
Integer[] nums1Array = nums1.toArray(new Integer[0]);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums1Array.length; i++) {
for (int j = i+1; j < nums1Array.length; j++) {
for (int k = j+1; k < nums1Array.length; k++) {
// mask = nums[i]+nums[j];
if((nums1Array[i]+nums1Array[j]+nums1Array[k])==0){
List<Integer> list = new ArrayList<>(Arrays.asList(nums1Array[i],nums1Array[j],nums1Array[k]));
Collections.sort(list);
result.add(list);
}
}
}
}
Set<List<Integer>> set = new HashSet<>(result);
List<List<Integer>> uniqueList = new ArrayList<>(set);
return uniqueList;
}
上面的这一个是优化失败的，我想着可以转化成集合去除里面重复的元素，就可以不用算这么多了，但是多余的元素在一定情况下也是有用的，如果你要去除还要进行判断。这个判断就很难进行下去，把数据清洗单独放到前面，确实没有什么好的方法做判断。
既然数据清洗放到上面不行，那我就放到下面的for里面去。但是传统的arrays类型很难进行增删操作，java给了arraylist可以进行方便的增删操作。我们就要先用java8的新特性流来转化数组变成list。然后我再进行判断，如果result中已经含有对应的list，说明这三个元素重复了，就可以进行安全删除。不过我写到一半发现，如果删了的话，for循环的边界就变得比较难以确定了。于是又放弃了。
public static List<List<Integer>> threeSum3(int[] nums) {
Arrays.sort(nums);
List<Integer> list = Arrays.stream(nums)
.boxed()
.collect(Collectors.toList());
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < list.size(); i++) {
for (int j = list.size()-1; j > i+1; j--) {
int find = Collections.binarySearch(list,(-list.get(i)-list.get(j)));
if (find!= -1&&find !=i&&find !=j){
List<Integer> list1 = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[find]));
Collections.sort(list1);
result.add(list1);
if (result.contains(list1)) {
list.remove(i);
list.remove(j);
list.remove(find);
}
}
}
}
Set<List<Integer>> set = new HashSet<>(result);
List<List<Integer>> uniqueList = new ArrayList<>(set);
return uniqueList;
}
这样的数据清洗思路似乎没办法，那我回到把数据清洗放到前面去，用哈希表先存储一遍数组，记录每一个数出现的次数。如果超过某个数那就不把它读到处理的数组中去。
public static List<List<Integer>> threeSum5(int[] nums) {
int k = 100; // 允许的最大重复次数
Map<Integer, Integer> countMap = new HashMap<>();
for (int num : nums) {
countMap.put(num, countMap.getOrDefault(num, 0) + 1);
}
int index = 0,cnt=0;
for (int num : nums) {
if (countMap.get(num) <= k) {
nums[index++] = num;
cnt++;
}
}
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
for (int j = nums.length-1; j > i+1; j--) {
int find = binarySearch(nums,i+1,j,(0-nums[i]-nums[j]));
if (find!= -1&&find !=i&&find !=j){
List<Integer> list = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[find]));
Collections.sort(list);
result.add(list);
}
}
}
Set<List<Integer>> set = new HashSet<>(result);
List<List<Integer>> uniqueList = new ArrayList<>(set);
return uniqueList;
}
这种其实只能防止一些特殊的测试样例，比如全是0的，但是如果别人刚好需要这么多个，你把别人都清洗掉，就会出现不对。
第四次优化
这一次gpt给了我一点思路，我如果要去掉重复元素的话，可以在处理完之后添加跳过语句，这样就不会影响程序的正常判断了。
public static List<List<Integer>> threeSum6(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
if (nums[i]>0)break;
for (int j = nums.length-1; j > i+1; j--) {
int find = binarySearch(nums,i+1,j,(0-nums[i]-nums[j]));
if (find!= -1&&find !=i&&find !=j){
List<Integer> list = new ArrayList<>(Arrays.asList(nums[i],nums[j],nums[find]));
Collections.sort(list);
result.add(list);
}
//必须放到后面不能放到if之前
while (j>i+3&&nums[j]==nums[j-1])j--;
}
//这个也是
while (i<nums.length-3&&nums[i]==nums[i+1])i++;
}
Set<List<Integer>> set = new HashSet<>(result);
List<List<Integer>> uniqueList = new ArrayList<>(set);
return uniqueList;
这一个解已经可以accept了，但是姿势十分扭曲，哈哈哈哈哈。
第五次优化
我看了一下别人的题解，用到了双指针、排序、跳过。只用了30ms，解和解的差距怎么这么大？人和人的差距这么大？
public static List<List<Integer>> threeSum7(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
if (nums[i] > 0) break;
if (i > 0 && nums[i] == nums[i - 1]) continue; // 避免重复计算
int left = i + 1, right = nums.length - 1;
while (left < right) {
int sum = nums[left] + nums[right] + nums[i];
if (sum == 0) {
result.add(Arrays.asList(nums[i], nums[left], nums[right]));
while (left < right && nums[left] == nums[left + 1]) left++; // 跳过重复元素
while (left < right && nums[right] == nums[right - 1]) right--; // 跳过重复元素
left++;
right--;
} else if (sum < 0) {
left++;
} else {
right--;
}
}
}
return new ArrayList<>(new LinkedHashSet<>(result));
}
第六次优化
这一个只用了20ms，超过了百分之九十的人，解和解的差距怎么这么大？人和人的差距这么大？
public static List<List<Integer>> threeSum8(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length - 2; i++) {
if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) { // 避免重复计算
int left = i + 1, right = nums.length - 1, target = -nums[i];
while (left < right) {
if (nums[left] + nums[right] == target) {
result.add(Arrays.asList(nums[i], nums[left], nums[right]));
while (left < right && nums[left] == nums[left + 1]) left++; // 跳过重复元素
while (left < right && nums[right] == nums[right - 1]) right--; // 跳过重复元素
left++;
right--;
} else if (nums[left] + nums[right] < target) {
left++;
} else {
right--;
}
}
}
}
return result;
}
第七次优化
这个是最强的解，没有之一，只是用了3ms,写得我都看不懂。
public List<List<Integer>> threeSum(int[] nums) {
return nSum(nums, 3, 0);
}
public List<List<Integer>> nSum(int[] nums, int k, int target) {
return new java.util.AbstractList<List<Integer>>() {
final List<List<Integer>> res = new ArrayList<>();
final List<Integer> path = new ArrayList<>();
long min;
public List<Integer> get(int index) {
init();
return res.get(index);
}
public int size() {
init();
return res.size();
}
public void init() {
if (res.isEmpty()) {
int n = nums.length;
long[] Arr = new long[n];
Arrays.sort(nums);
min = nums[0];
for (int i = 0; i < n; i++) {
Arr[i] = nums[i] - min;
}
long NewTarget = (long) target - (long) k * min;
C(false, Arr, n, k, NewTarget);
}
}
//C(n, k) = C(n - 1, k) + C(n - 1, k - 1)
public void C(boolean T, long[] a, int n, int k, long target) {
if (n == 0 || k == 0) {
if (target == 0 && k == 0) {
res.add(new ArrayList<>(path));
}
return;
}
if (k == 2) {
if (!T && n != a.length && a[n] == a[n - 1]) {
return;
}
//两数之和模板
twoSum(a, 0, n - 1, target);
return;
}
if (n == k) {
if (!T && n != a.length && a[n] == a[n - 1]) {
return;
}
//数组中元素和是否等于target
sumArr(a, n, target);
return;
}
if (check(a, n, k, target)) {
return;
}
C(false, a, n - 1, k, target);
if (!T && n != a.length && a[n] == a[n - 1]) {
return;
}
if (target - a[n - 1] >= 0) {
path.add((int) (a[n - 1] + min));
C(true, a, n - 1, k - 1, target - a[n - 1]);
path.remove(path.size() - 1);
}
}
void twoSum(long[] a, int l, int r, long target) {
if (l >= r || a[r - 1] + a[r] < target || a[l] + a[l + 1] > target) {
return;
}
while (r > l) {
long sum = a[l] + a[r];
if (sum < target) {
l++;
} else if (sum > target) {
r--;
} else {
path.add((int) (a[l] + min));
path.add((int) (a[r] + min));
res.add(new ArrayList<>(path));
path.remove(path.size() - 1);
path.remove(path.size() - 1);
while (r > l && a[l] == a[l + 1]) {
l++;
}
while (r > l && a[r] == a[r - 1]) {
r--;
}
l++;
r--;
}
}
}
void sumArr(long[] a, int n, long target) {
for (int i = n - 1; i > -1; i--) {
target -= a[i];
path.add((int) (a[i] + min));
}
if (target == 0) {
res.add(new ArrayList<>(path));
}
for (int i = n - 1; i > -1; i--) {
target += a[i];
path.remove(path.size() - 1);
}
}
boolean check(long[] a, int n, int k, long target) {
if (n - k < 0) {
return true;
}
long max = 0;
long min = 0;
for (int i = 0; i < k; i++) {
min += a[i];
max += a[n - i - 1];
}
if (target < min || target > max) {
return true;
}
return false;
}
};
}
