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题目

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.

分析

• 最直观的当然是递归了
• 因为是两个变量，二维矩阵是少不了的
• d[i][j]可以表达，当s1取前i个，s2取前j个，是否可以组成s3取前i+j个？
• 编程上面有技巧，不能让d[s1.length][s2.length]，这样index只能取到s1.length-1和s2.length-1
• 推导公式，关键就在s1[i-1]==s3[i+j-1]这一句，因为是取s1的前i个，那最后一个就是s[i-1]了，这一点要严格按照d[i][j]的定义来

d[i][j] = (d[i-1][j]&&s1[i -1 ]==s3[i+j -1 ]) || (d[i][j-1]&&s2[j-1]==s3[i+j-1])

代码

package InterleaveString;

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {

        if (s1.length() + s2.length() != s3.length()) {
            return false;
        }

        boolean[][] d = new boolean[s1.length()+1][s2.length()+1];
        d[0][0] = true;

        for (int i = 1; i <= s1.length(); i++) {
            if(s1.charAt(i-1) == s3.charAt(i-1)) {
                d[i][0] = true;
            } else {
                break;
            }
        }

        for (int j = 1; j <= s2.length(); j++) {
            if (s2.charAt(j-1) == s3.charAt(j-1)) {
                d[0][j] = true;
            } else {
                break;
            }
        }

        for (int i = 1; i <= s1.length(); i++) {
            for (int j = 1; j <= s2.length(); j++) {
                d[i][j] |= d[i - 1][j] && s1.charAt(i-1) == s3.charAt(i + j-1);
                d[i][j] |= d[i][j - 1] && s2.charAt(j-1) == s3.charAt(i + j-1);
            }
        }


        return d[s1.length()][s2.length()];
    }

}

Interleaving String

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原文地址：http://my.oschina.net/zuoyc/blog/347676