Interleaving String，interleavingstring

Interleaving String，interleavingstring

Given s1, s2, s3, find whethers3 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.

最簡單思路就是遞迴去做，但是這樣需要遞迴最多s3長度的層次，消耗比較大，很容易出現逾時錯誤，進行遞迴嘗試果然出現逾時錯誤，但是遞迴是解決這個問題最直觀的方法。

遞迴方法不行考慮DP，我們用path[i][j] 記錄s1到i 和s2到j 和s3[i+j-1]是否滿足要求，這個過程就是填一張二維數組表的過程，先初始化第一行和第一列，然後填表計算出所有，返回path[s1.length][s2.length]的值。

已經AC的代碼：

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {if(s1.length()+s2.length()!=s3.length()){return false;}int row = s1.length();int col = s2.length();boolean [][]path = new boolean[row+1][col+1];path[0][0] = true;for(int i=1;i<=row;i++){path[i][0] = path[i-1][0] && (s1.charAt(i-1)==s3.charAt(i-1));}for(int i=1;i<=col;i++){path[0][i] = path[0][i-1] && (s2.charAt(i-1) == s3.charAt(i-1));}for(int i=1;i<=row;i++){for(int j=1;j<=col;j++){path[i][j] = (path[i-1][j] && s1.charAt(i-1)==s3.charAt(i+j-1)) ||(path[i][j-1] && s2.charAt(j-1) == s3.charAt(i+j-1));}}return path[row][col];}}