97. Interleaving String
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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.
题意:给定3个字符串s1,s2,s3。证明s3是否由s1和s2互相插入所形成;
一看题目就有了思路,难度标注的居然是Hard,很开心,于是开始敲代码:
思路:使用递归,每次都判断s3的首字符是否等于s1和s2的首字符
1、s3首字符等于s1首字符,不等于s2首字符,递归调用方法 isInterleave(s1.substring(1),s2,s3.substring(1))
2、s3首字符等于s2首字符,不等于s1首字符,递归调用方法 isInterleave(s1,s2.substring(1),s3.substring(1))
3、s3首字符等于s2首字符,同时等于s1首字符,递归调用方法 isInterleave(s1.substring(1),s2,s3.substring(1)) 和isInterleave(s1,s2.substring(1),s3.substring(1)),二者只要有一个为true,结果为true;
public class Solution { public boolean isInterleave(String s1, String s2, String s3) { if(s1.length()+s2.length() != s3.length()) return false;//长度不等直接返回false if(s1.length()==0){ //考虑s1或者s2为空的特殊情况 if(s2.equals(s3)) return true; else return false; } if(s2.length()==0){ if(s1.equals(s3)) return true; else return false; } if(s3.charAt(0)==s1.charAt(0) || s3.charAt(0)==s2.charAt(0)){ if(s3.charAt(0)==s1.charAt(0) && s3.charAt(0)==s2.charAt(0)){ //s3和s1,s2的首字符相同,任意匹配其中一种即可 return (isInterleave(s1.substring(1),s2,s3.substring(1)) || isInterleave(s1,s2.substring(1),s3.substring(1))); }else if(s3.charAt(0)==s1.charAt(0)){ //s3和s1的首字符相同 return isInterleave(s1.substring(1),s2,s3.substring(1)); }else if(s3.charAt(0)==s2.charAt(0)){ //s3和s2的首字符相同 return isInterleave(s1,s2.substring(1),s3.substring(1)); } } return false; }}
@以下代码来自leetcode,
public boolean isInterleave(String s1, String s2, String s3) { if ((s1.length()+s2.length())!=s3.length()) return false; boolean[][] matrix = new boolean[s2.length()+1][s1.length()+1]; matrix[0][0] = true; for (int i = 1; i < matrix[0].length; i++){ matrix[0][i] = matrix[0][i-1]&&(s1.charAt(i-1)==s3.charAt(i-1)); } for (int i = 1; i < matrix.length; i++){ matrix[i][0] = matrix[i-1][0]&&(s2.charAt(i-1)==s3.charAt(i-1)); } for (int i = 1; i < matrix.length; i++){ for (int j = 1; j < matrix[0].length; j++){ matrix[i][j] = (matrix[i-1][j]&&(s2.charAt(i-1)==s3.charAt(i+j-1))) || (matrix[i][j-1]&&(s1.charAt(j-1)==s3.charAt(i+j-1))); } } return matrix[s2.length()][s1.length()];}数组matrix[i][j]表示: s2的前i个字符组成的字符串与s1的前j个字符组成的字符串 是否可以通过互相插入组成 s3的前i+j个字符组成的字符串;
状态转移方程式:
matrix[i][j] = (matrix[i-1][j]&&(s2.charAt(i-1)==s3.charAt(i+j-1)) || (matrix[i][j-1]&&(s1.charAt(j-1)==s3.charAt(i+j-1)));
又matrix[i][0]与matrix[0][j]容易求得,故可通过循环得到matrix[s2.length()][s1.length()]的值,从而解决此题。
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