97. Interleaving String-优快云博客

本文介绍了一种使用动态规划解决字符串交织问题的方法。通过构建二维DP表，判断字符串s3是否能由s1和s2交织而成。文章详细阐述了DP表的初始化及状态转移方程。

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

For example,
Given:
s1 = “aabcc”,
s2 = “dbbca”,

When s3 = “aadbbcbcac”, return true.
When s3 = “aadbbbaccc”, return false.
解决思路
使用动态规划，dp[i][j] 表示 s1[i-1] 和 s2[j-1]能否组合成s3[i+j-1]，若可以，则true，否则false。
初始情况：dp[0][0] = true , dp[i][0] = s1[i-1] == s3[i-1] ? dp[i-1][0] : 0 ,
dp[0][i] = s2[i-1] == s3[i-1] ? dp[0][i-1] : 0
状态转移：
if (s3[i+j-1] == s1[i-1])
dp[i][j] = dp[i-1][j];
if (s3[i+j-1] == s2[j-1])
dp[i][j] = dp[i][j-1] || dp[i][j];
代码

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