构造字典序最大的合并字符串【LC1754】

You are given two strings word1 and word2. You want to construct a string merge in the following way: while either word1 or word2 are non-empty, choose one of the following options:

• If word1 is non-empty, append the first character in word1 to mergeand delete it from word1
  • For example, if word1 = "abc" and merge = "dv", then after choosing this operation, word1 = "bc" and merge = "dva".
• If word2is non-empty, append the first character in word2 to merge and delete it from word2
  • For example, if word2 = "abc" and merge = "", then after choosing this operation, word2 = "bc" and merge = "a".

虽然烤红薯很好吃 但是被烤箱烫伤了瞬间就不香了
现在的医院全是咳嗽的哎 真可怕哎

• 思路：双指针遍历两个字符串，贪心比较字符的字典顺序，并添加至结果集

  局部最优：优先选择后缀字符串的字典顺序较大的字符至结果集

  全局最优：合并后的字符串的字典顺序最大

  • 如果两个指针对应的字符不相等，直接将字典顺序较大的字符加入结果集的末尾，并且移动改字符串对应的指针
  • 如果两个指针对应的字符相等，那么需要比较后缀字符串的字典顺序
    • 添加较大者对应的字符至结果集
    • 如果字典顺序相同，那么添加任意字符均不会影响结果

• 实现

  • 先使用双指针对字符进行比较直至移动到一个字符串的末尾
  • 此时直接将另一个非空字符串剩余的字符添加至结果集即可

  class Solution {
      public String largestMerge(String word1, String word2) {
          StringBuilder sb = new StringBuilder();
          int n = word1.length(), m = word2.length();
          int i = 0, j = 0;
          while (i < n && j < m){
              char ch1 = word1.charAt(i), ch2 = word2.charAt(j);
              if (ch1 > ch2){
                  sb.append(ch1);
                  i++;
              }else if (ch1 < ch2){
                  sb.append(ch2);
                  j++;
              }else{
                  // 字符相等 需要判断后续字符的字典顺序
                  if (i < n && word1.substring(i).compareTo(word2.substring(j)) > 0){
                      sb.append(ch1);
                      i++;
                  }else{
                      sb.append(ch2);
                      j++;
                  }
              }      
          }
          if (i < n){
              sb.append(word1.substring(i));
          }
          if (j < m){
              sb.append(word2.substring(j));
          }
          return new String(sb);
      }
  }
  

  • 复杂度

    

    • 时间复杂度：$O(min(m,n)\ast min(m,n))$，$m$和$n$分别为字符串的长度，先压入字符进行比较，需要比较$min(m,n)$次，若两个字符串相等那么需要进行后缀比较的时间复杂度为$O(min(m,n))$次
    • 空间复杂度：$O(m+n)$，生成的后缀字符串所需要的空间为$O(m+n)$

• 实现：每一次比较均比较后缀字符串

  代码好看，时间复杂度稍微高一点

  class Solution {
      public String largestMerge(String word1, String word2) {
          StringBuilder merge = new StringBuilder();
          int i = 0, j = 0;
          while (i < word1.length() || j < word2.length()) {
              if (i < word1.length() && word1.substring(i).compareTo(word2.substring(j)) > 0) {
                  merge.append(word1.charAt(i));
                  i++;
              } else {
                  merge.append(word2.charAt(j));
                  j++;
              }
          }
          return merge.toString();
      }
  }
  
  作者：力扣官方题解
  链接：https://leetcode.cn/problems/largest-merge-of-two-strings/solutions/2030226/gou-zao-zi-dian-xu-zui-da-de-he-bing-zi-g6az1/
  来源：力扣（LeetCode）
  著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
  

  • 复杂度

    

    • 时间复杂度：$O((m+n)\ast max(m,n))$，$m$和$n$分别为字符串的长度，先压入字符进行比较，需要比较$(m+n)$次，每次需要进行后缀比较的时间复杂度为$O(max(m,n))$次
    • 空间复杂度：$O(m+n)$，生成的后缀字符串所需要的空间为$O(m+n)$