【LetMeFly】132.分割回文串 II：动态规划
力扣题目链接：https://leetcode.cn/problems/palindrome-partitioning-ii/
给你一个字符串 s，请你将 s 分割成一些子串，使每个子串都是回文串。
返回符合要求的 最少分割次数 。
示例 1：
输入：s = "aab"
输出：1
解释：只需一次分割就可将 s 分割成 ["aa","b"] 这样两个回文子串。示例 2：
输入：s = "a"
输出：0示例 3：
输入：s = "ab"
输出：1提示：
1 <= s.length <= 2000s仅由小写英文字母组成
解题方法：动态规划
整个过程分为两步：预处理 和 动态规划
动态规划：
使用数组 d p dp dp，其中 d p [ i ] dp[i] dp[i]代表使得子字符串 0... i 0...i 0...i为回文字符串组合的最小分割次数，那么 d p [ l e n ( s ) − 1 ] dp[len(s) - 1] dp[len(s)−1]即为答案。
-
如果 0... i 0...i 0...i直接为回文字符串，那么分割次数为0。
-
否则，对于 j ∈ 0... i − 1 j\in 0...i-1 j∈0...i−1，如果 j + 1.. i j + 1..i j+1..i是回文字符串，那么有 d p [ i ] = m i n ( d p [ j ] + 1 ) dp[i] = min(dp[j] + 1) dp[i]=min(dp[j]+1)
预处理：
有没有什么办法 O ( 1 ) O(1) O(1)时间内快速判断下标从 i i i到 j j j的子字符串是否为回文字符串？有，我们可以先使用 O ( n 2 ) O(n^2) O(n2)复杂度的时间预处理。使用 i s O k [ i ] [ j ] isOk[i][j] isOk[i][j]表示子字符串 i . . . j i...j i...j是否为回文字符串：
- 如果子字符串为空或者长度为1，则是回文字符串( i ≥ j i \geq j i≥j)
- 否则：是回文字符串当且仅当 s [ i ] = = s [ j ] AND i s O k [ i + 1 ] [ j − 1 ] s[i] == s[j] \text{ AND }isOk[i + 1][j - 1] s[i]==s[j] AND isOk[i+1][j−1]
时空复杂度分析
- 时间复杂度 O ( n 2 ) O(n^2) O(n2)，预处理和动态规划的时间复杂度都是 O ( n 2 ) O(n^2) O(n2)。其中 n = l e n ( s ) n = len(s) n=len(s)
- 空间复杂度 O ( n 2 ) O(n^2) O(n2)
AC代码
C++
cpp
/*
* @Author: LetMeFly
* @Date: 2025-03-02 12:02:45
* @LastEditors: LetMeFly.xyz
* @LastEditTime: 2025-03-02 12:26:06
*/
class Solution {
public:
int minCut(string s) {
vector<vector<bool>> isOk(s.size(), vector<bool>(s.size(), true));
for (int i = s.size() - 1; i >= 0; i--) {
for (int j = i + 1; j < s.size(); j++) {
isOk[i][j] = s[i] == s[j] && isOk[i + 1][j - 1];
}
}
vector<int> dp(s.size(), 1000000);
for (int i = 0; i < s.size(); i++) {
if (isOk[0][i]) {
dp[i] = 0;
continue;
}
for (int j = 0; j < i; j++) {
if (isOk[j + 1][i]) {
dp[i] = min(dp[i], dp[j] + 1);
}
}
}
return dp.back();
}
};Python
python
'''
Author: LetMeFly
Date: 2025-03-02 12:26:57
LastEditors: LetMeFly.xyz
LastEditTime: 2025-03-02 12:33:40
'''
class Solution:
def minCut(self, s: str) -> int:
isOk = [[True] * len(s) for _ in range(len(s))]
for i in range(len(s) - 1, -1, -1):
for j in range(i + 1, len(s)):
isOk[i][j] = s[i] == s[j] and isOk[i + 1][j - 1]
dp = [100000] * len(s)
for i in range(len(s)):
if isOk[0][i]:
dp[i] = 0
continue
for j in range(i):
if isOk[j + 1][i]:
dp[i] = min(dp[i], dp[j] + 1)
return dp[-1]Java
java
/*
* @Author: LetMeFly
* @Date: 2025-03-02 12:34:31
* @LastEditors: LetMeFly.xyz
* @LastEditTime: 2025-03-02 12:38:17
*/
class Solution {
public int minCut(String s) {
boolean[][] isOk = new boolean[s.length()][s.length()];
for (int i = 0; i < s.length(); i++) {
for (int j = 0; j < s.length(); j++) {
isOk[i][j] = true;
}
}
for (int i = s.length() - 1; i >= 0; i--) {
for (int j = i + 1; j < s.length(); j++) {
isOk[i][j] = s.charAt(i) == s.charAt(j) && isOk[i + 1][j - 1];
}
}
int[] dp = new int[s.length()];
for (int i = 0; i < s.length(); i++) {
dp[i] = 100000;
}
for (int i = 0; i < s.length(); i++) {
if (isOk[0][i]) {
dp[i] = 0;
continue;
}
for (int j = 0; j < i; j++) {
if (isOk[j + 1][i]) {
dp[i] = Math.min(dp[i], dp[j] + 1);
}
}
}
return dp[dp.length - 1];
}
}Go
go
/*
* @Author: LetMeFly
* @Date: 2025-03-02 12:39:13
* @LastEditors: LetMeFly.xyz
* @LastEditTime: 2025-03-02 12:43:12
*/
package main
func minCut(s string) int {
isOk := make([][]bool, len(s))
for i, _ := range isOk {
isOk[i] = make([]bool, len(s))
for j, _ := range isOk[i] {
isOk[i][j] = true
}
}
for i := len(s) - 1; i >= 0; i-- {
for j := i + 1; j < len(s); j++ {
isOk[i][j] = s[i] == s[j] && isOk[i + 1][j - 1]
}
}
dp := make([]int, len(s))
for i, _ := range dp {
dp[i] = 100000
}
for i := 0; i < len(dp); i++ {
if isOk[0][i] {
dp[i] = 0
continue
}
for j := 0; j < i; j++ {
if isOk[j + 1][i] {
dp[i] = min(dp[i], dp[j] + 1)
}
}
}
return dp[len(dp) - 1]
}同步发文于CSDN和我的个人博客，原创不易，转载经作者同意后请附上原文链接哦~
千篇源码题解已开源