[leetcode]97. Interleaving String-优快云博客

本文链接：https://blog.youkuaiyun.com/xiaocong1990/article/details/80482137

本文探讨了如何判断一个字符串是否可以通过两个给定字符串交织而成的问题。提供了两种解决方案：一种使用动态规划，另一种采用贪心策略。动态规划方法通过构建二维布尔矩阵来记录子问题的解，而贪心策略则试图尽可能匹配第一个字符串，遇到不匹配的情况再尝试匹配第二个字符串。

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链接：https://leetcode.com/problems/interleaving-string/description/

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

Example 1:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true

Example 2:

Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

动态规划

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        if (s1.length() + s2.length() != s3.length()) return false;
        vector<vector<bool>> dp(s1.length() + 1, vector<bool>(s2.length() + 1, false));
        dp[0][0] = true;
        
        for(int i=1; i<=s1.length(); i++)  if (s3.substr(0, i) == s1.substr(0, i)) dp[i][0] = true;
        for(int j=1; j<=s2.length(); j++) if (s3.substr(0, j) == s2.substr(0, j)) dp[0][j] = true;
        for(int i=0; i<s1.length(); i++) {
            for(int j=0; j<s2.length(); j++) {
                if (s3[i + j + 1] != s1[i] && s3[i + j + 1] != s2[j]) dp[i + 1][j + 1] = false;
                if (s3[i + j + 1] == s1[i]) dp[i + 1][j + 1] = dp[i + 1][j + 1] || dp[i][j + 1];
                if (s3[i + j + 1] == s2[j]) dp[i + 1][j + 1] = dp[i + 1][j + 1] || dp[i + 1][j];
            }
        }
        return dp.back().back();
    }
};

贪心

Travel s3 to find s1 greedily, if meet s3[k] != s1[i], we try to check whether s3[k] == s2[j], if no, go backward k and i, incease c, set c as the backward steps count.

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        int i=0,j=0,k=0,c = 0;
        while(k < s3.length()) {
            if(!c && s3[k] == s1[i]) { i++; k++; continue; }
            if(s3[k] == s2[j]) { j++; k++; c = 0; continue; }
            --i; --k;    c++;
            if(k < 0) break;
        }
        return k == s3.length() && i == s1.length() && j == s2.length();
    }
};