[leetcode]Interleaving String

一开始用递归。大数据超时了。后来网上看了一下，是动态规划。归根结底就是有重复子问题，比如bbbbaaaa由bbaa和bbaa来匹配的话，那么无论第一个b选s1还是s2的，都会出现baa和baa这样的子问题。

那么由于s1+s2==s3，只需要用二维数组记录状态就行了。状态转移方程可详见：http://blog.unieagle.net/2012/09/29/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Ainterleaving-string%EF%BC%8C%E4%BA%8C%E7%BB%B4%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/

这是最后过得动态规划版本：（我看有的人把数组起名为matrix）

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if (s1.length() + s2.length() != s3.length()) return false;
        boolean[][] status = new boolean[s1.length()+1][s2.length()+1];
        
        int len1 = s1.length();
        int len2 = s2.length();
        int len3 = s3.length();
        
        status[0][0] = true;
        for (int i = 1; i < s1.length()+1; i++) {
            status[i][0] = status[i-1][0] && s1.charAt(len1 - i) == s3.charAt(len3 - i);
        }
        
        for (int j = 1; j < s2.length()+1; j++) {
            status[0][j] = status[0][j-1] && s2.charAt(len2 - j) == s3.charAt(len3 - j);
        }
        
        for (int i = 1; i < s1.length()+1; i++) {
            for (int j = 1; j < s2.length()+1; j++) {
                status[i][j] = s3.charAt(len3 - i - j) == s1.charAt(len1 - i) && status[i-1][j] ||
                               s3.charAt(len3 - i - j) == s2.charAt(len2 - j) && status[i][j-1];
                
            }
        }
        
        return status[len1][len2];
    }
}

这是大数据超时的版本，我想如果用动态规划那个数组记录一下计算过的状态也是能过得。

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if (s1.length() + s2.length() != s3.length()) return false;
        return isInterleave(s1, s2, s3, 0, 0, 0);
        
    }
    
    public boolean isInterleave(String s1, String s2, String s3, int idx1, int idx2, int idx3) {
        if (idx3 == s3.length()) {
            if (idx1 == s1.length() && idx2 == s2.length()) return true;
            else return false;
        }
        
        boolean cond1 = idx1 < s1.length() && s1.charAt(idx1) == s3.charAt(idx3);
        boolean cond2 = idx2 < s2.length() && s2.charAt(idx2) == s3.charAt(idx3);
        if (cond1 && cond2) {
            return isInterleave(s1, s2, s3, idx1+1, idx2, idx3+1) ||
                    isInterleave(s1, s2, s3, idx1, idx2+1, idx3+1);
        }
        else if (cond1) {
            return isInterleave(s1, s2, s3, idx1+1, idx2, idx3+1);
        }
        else if (cond2) {
            return isInterleave(s1, s2, s3, idx1, idx2+1, idx3+1);
        }
        return false;
    }
    
}