Interleaving String
https://leetcode.com/problems/interleaving-string/
给定三个字符串,s1,s2,s3,看s3是否由s1和s2穿插组成
Example 1:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Example 2:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
解法一 超时啊,超时啊
public boolean isInterleave(String s1, String s2, String s3) {
if (s1.isEmpty() && s2.isEmpty() && s3.isEmpty()) {
return true;
}
//s3 为空,但是其他两个不为空,返回false
if(s3.isEmpty()){
return false;
}
if (!s1.isEmpty() && s1.charAt(0) == s3.charAt(0)) {
if (isInterleave(s1.substring(1), s2, s3.substring(1))) {
return true;
}
}
if (!s2.isEmpty() && s2.charAt(0) == s3.charAt(0)) {
if (isInterleave(s1, s2.substring(1), s3.substring(1))) {
return true;
}
}
return false;
}
解法二 跟解法一原理一样
多一个数组,去判断,每一位上已经判断过的条件,避免重复判断
public boolean isInterleave(String s1, String s2, String s3) {
int memo[][] = new int[s1.length()][s2.length()];
for (int i = 0; i < s1.length(); i++) {
for (int j = 0; j < s2.length(); j++) {
memo[i][j] = -1;
}
}
return is_Interleave(s1, 0, s2, 0, s3, 0, memo);
}
public boolean is_Interleave(String s1, int i, String s2, int j, String s3, int k, int[][] memo) {
if (i == s1.length()) {
return s2.substring(j).equals(s3.substring(k));
}
if (j == s2.length()) {
return s1.substring(i).equals(s3.substring(k));
}
if (memo[i][j] >= 0) {
return memo[i][j] == 1 ? true : false;
}
boolean ans = false;
if (s3.charAt(k) == s1.charAt(i) && is_Interleave(s1, i + 1, s2, j, s3, k + 1, memo)
|| s3.charAt(k) == s2.charAt(j) && is_Interleave(s1, i, s2, j + 1, s3, k + 1, memo)) {
ans = true;
}
memo[i][j] = ans ? 1 : 0;
return ans;
}
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