97. 交错字符串

 

 方法一： 动态规划

class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        int n1 = s1.length(), n2 = s2.length(), n3 = s3.length();
        if(n1 + n2 != n3) return false;
        char[] arr1 = s1.toCharArray(), arr2 = s2.toCharArray(), arr3 = s3.toCharArray();
        boolean[][] dp = new boolean[n1+1][n2+1];
        dp[0][0] = true;
        for(int i = 1; i <= n1; i++) {
            if(dp[i-1][0] && arr1[i-1] == arr3[i-1]) dp[i][0] = true;
        }
        for(int i = 1; i <= n2; i++) {
            if(dp[0][i-1] && arr2[i-1] == arr3[i-1]) dp[0][i] = true;
        }
        for(int i = 1; i <= n1; i++) {
            for(int j = 1; j <= n2; j++) {
                if(arr1[i-1] == arr3[i+j-1]) {
                    dp[i][j] = dp[i-1][j];
                }
                if(arr2[j-1] == arr3[i+j-1]) {
                    dp[i][j] = dp[i][j] || dp[i][j-1];
                }
            }
        }
        return dp[n1][n2];
    }
}

 方法二：记忆化回溯

class Solution {
    char[] arr1;
    char[] arr2;
    char[] arr3;
    public boolean isInterleave(String s1, String s2, String s3) {
        int n1 = s1.length(), n2 = s2.length(), n3 = s3.length();
        if(n1 + n2 != n3) return false;
        visited = new boolean[n1+1][n2+1];
        arr1 = s1.toCharArray();
        arr2 = s2.toCharArray();
        arr3 = s3.toCharArray();
        return dfs(0,0,0);
    }
    boolean[][] visited;
    public boolean dfs(int i, int j, int k) {
        if(k == arr3.length) return true;
        if(visited[i][j]) return false;
        else if(i < arr1.length && arr1[i] == arr3[k] && dfs(i+1,j,k+1)) return true;
        else if(j < arr2.length && arr2[j] == arr3[k] && dfs(i,j+1,k+1)) return true;
        visited[i][j] = true;
        return false;
    }
}