Interleaving String leetcode java
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.
题解:
这道题还是像之前我引过的那句话:
“When you see string problem that is about subsequence or matching,
dynamic programming method should come to your mind naturally. ”
所以这道题还是用DP的思想解决。
大体思路是,s1取一部分s2取一部分,最后是否能匹配s3。
动态规划数组是dp[i][j],表示:s1取前i位,s2取前j位,是否能组成s3的前i+j位。
初始化是,假设s1为空,那么s2每一位跟s3匹配放入dp[0][j];假设s2为空,那么s1每一位跟s3匹配放入dp[i][0]。
下面就继续匹配。讲解引用自: http://blog.csdn.net/u011095253/article/details/9248073
“
那什么时候取True,什么时候取False呢?
False很直观,如果不等就是False了嘛。
那True呢?首先第一个条件,新添加的字符,要等于s3里面对应的位( i + j 位),第二个条件,之前那个格子也要等于True
举个简单的例子s1 = ab, s2 = c, s3 = bbc ,假设s1已经取了2位,c还没取,此时是False(ab!=bb),我们取s2的新的一位c,即便和s3中的c相等,但是之前是False,所以这一位也是False
同理,如果s1 = ab, s2 = c, s3=abc ,同样的假设,s1取了2位,c还没取,此时是True(ab==ab),我们取s2的新的一位c,和s3中的c相等,且之前这一位就是True,此时我们可以放心置True (abc==abc)
”
代码如下:
1 public boolean isInterleave(String s1, String s2, String s3) {
2 if(s3.length()!=s1.length()+s2.length())
3 return false;
4
5 boolean [][] dp = new boolean [s1.length()+1][s2.length()+1];
6 dp[0][0]=true;
7
8 for(int i = 1; i<=s1.length() && s1.charAt(i-1)==s3.charAt(i-1); i++)
9 dp[i][0]=true;
for(int i = 1; i<=s2.length() && s2.charAt(i-1)==s3.charAt(i-1); i++)
dp[0][i]=true;
for(int i = 1; i <= s1.length(); i++){
for(int j = 1; j <= s2.length(); j++){
char c = s3.charAt(i+j-1);
if(c == s1.charAt(i-1) && dp[i-1][j])
dp[i][j] = true;
if(c == s2.charAt(j-1) && dp[i][j-1])
dp[i][j] = true;
}
}
return dp[s1.length()][s2.length()];
}
Reference:
http://blog.csdn.net/u011095253/article/details/9248073
http://www.cnblogs.com/lichen782/p/leetcode_interleaving_string.html
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