本系列文章记录labuladong的算法小抄中剑指offer题目

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剑指 Offer II 018. 有效的回⽂

题目链接：https://leetcode.cn/problems/XltzEq/

class Solution {
    public boolean isPalindrome(String s) {
        // 先把所有字符转化成小写，并过滤掉空格和标点这类字符
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (Character.isLetterOrDigit(c)) {
                sb.append(Character.toLowerCase(c));
            }
        }

        // 然后对剩下的这些目标字符执行双指针算法，判断回文串
        s = sb.toString();
        int left = 0;
        int right = s.length()-1;
        while(right>left){
            if(s.charAt(left)!=s.charAt(right)){
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

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剑指 Offer 58 - I. 翻转单词顺序

题目链接：https://leetcode.cn/problems/fan-zhuan-dan-ci-shun-xu-lcof/

//先反转整个字符串，再逐个单词反转
class Solution {
    public String reverseWords(String s) {
        StringBuilder sb = trimSpaces(s);

        //整个字符串反转
        sb = sb.reverse();

        //逐个单词反转
        return reverseWord(sb);
    }

    //去除多余空格
    public StringBuilder trimSpaces(String s) {
        int left = 0, right = s.length() - 1;
        // 去掉字符串开头的空白字符
        while (left <= right && s.charAt(left) == ' ') {
            left++;
        }

        // 去掉字符串末尾的空白字符
        while (left <= right && s.charAt(right) == ' ') {
            right--;
        }

        // 将字符串间多余的空白字符去除
        StringBuilder sb = new StringBuilder();
        while(left<=right){
            char c=s.charAt(left);
            if(c!=' '){
                sb.append(c);
            }else if (sb.charAt(sb.length() - 1) != ' ') {
                sb.append(c);
            }
            left++;
        }
        return sb;
    }

    //逐单词反转
    public String reverseWord(StringBuilder sb){
        int left=0,right=0;
        while(right<sb.length()){
            while(right<sb.length() && sb.charAt(right)!=' '){right++;}
            //反转[left,right)的字符
            reverse(sb,left,right);
            right++;
            left=right;
        }
        return sb.toString();
    }

    //原地反转
    public void reverse(StringBuilder sb, int left, int right){
        while(right>left){
            char temp=sb.charAt(--right);
            sb.setCharAt(right,sb.charAt(left));
            sb.setCharAt(left,temp);
            left++;
        }
    }
}

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剑指 Offer 21. 调整数组顺序使奇数位于偶数前⾯

题目链接：https://leetcode.cn/problems/diao-zheng-shu-zu-shun-xu-shi-qi-shu-wei-yu-ou-shu-qian-mian-lcof/

//维护 nums前半部分 都是奇数
class Solution {
    public int[] exchange(int[] nums) {
        int left=0,right=0;
        while(right<nums.length){
            if(nums[right]%2==1){
                int temp=nums[right];
                nums[right]=nums[left];
                nums[left]=temp;
                left++;
            }
            right++;
        }
        return nums;
    }
}

//将后半部分奇数换到前半部分
class Solution {
    public int[] exchange(int[] nums) {
        int left=0,right=nums.length-1;
        while(left<right){
            while(left<right && nums[left]%2==1){
                left++;
            }
            int temp = nums[left];
            while(left<right && nums[right]%2==0){
                right--;
            }
            nums[left]=nums[right];
            nums[right]=temp;
        }
        return nums;
    }
}

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剑指 Offer 57. 和为s的两个数字

题目链接：https://leetcode.cn/problems/he-wei-sde-liang-ge-shu-zi-lcof/

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int left=0,right=nums.length-1;
        int[] res = new int[2];
        while(left<=right){
            if(nums[left]+nums[right]>target){
                right--;
            }else if(nums[left]+nums[right]<target){
                left++;
            }else{
                res[0]=nums[left];
                res[1]=nums[right];
                return res;
            }
        }
        return res;
    }
}

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剑指 Offer II 007. 数组中和为 0 的三个数

题目链接：https://leetcode.cn/problems/1fGaJU/

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        for(int i=0;i<nums.length-2;i++){
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int j=i+1;
            int k=nums.length-1;
            while(j<k){
                if(nums[i]+nums[j]+nums[k]<0){
                    j++;
                }else if(nums[i]+nums[j]+nums[k]>0){
                    k--;
                }else{
                    List<Integer> temp = new ArrayList<>();
                    temp.add(nums[i]);
                    temp.add(nums[j]);
                    temp.add(nums[k]);
                    res.add(temp);
                    while(j<k && nums[j]==nums[j+1]){j++;}
                    while(j<k && nums[j]==nums[k-1]){k--;}
                    j++;
                    k--;
                }
            }
        }
        return res;
    }
}

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总结

使用的内置函数：

//判断字符是否是字母和数字
Character.isLetterOrDigit(c)
//字符转化成小写
Character.toLowerCase(c)
//反转字符串
stringbuilder.reverse()