文章目錄

• • 1. 題目
  • 2. 解題

1. 題目

給定一個(gè)字符串 s，將 s 分割成一些子串，使每個(gè)子串都是回文串。

返回符合要求的最少分割次數(shù)。

示例: 輸入: "aab" 輸出: 1 解釋: 進(jìn)行一次分割就可將 s 分割成 ["aa","b"] 這樣兩個(gè)回文子串。

來(lái)源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/palindrome-partitioning-ii
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2. 解題

• dp[i]表示到 i 為止的子串最少需要分割多少次
• 如果一個(gè)子串為回文串，dp[i] = 0
• 如果不是，遍歷所有的 j （j <= i）,如果s[j,i]是回文串，dp[i] = min(dp[i], dp[j-1]+1)

28 / 29 個(gè)通過(guò)測(cè)試用例

# 超時(shí)例子 "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa" class Solution { public:int minCut(string s) {int i,j,n = s.size();vector<int> dp(n,0);if(s.size()<=1)return 0;for(i = 0; i < n; ++i)dp[i] = i;for(i = 1; i < n; ++i){for(j = i; j > 0; --j){if(ispalindrome(s,0,i))dp[i] = 0;else if(ispalindrome(s, j, i))dp[i] = min(dp[i], dp[j-1]+1);}}return dp[n-1];}bool ispalindrome(string& s, int l, int r){while(l < r){if(s[l++]!=s[r--])return false;}return true;} };

• 預(yù)先預(yù)處理得到所有可能的區(qū)間是否是是回文串
• 參考：LeetCode 5. 最長(zhǎng)回文子串（動(dòng)態(tài)規(guī)劃）

class Solution { public:int minCut(string s) {int i,j,len,n = s.size();vector<int> dp(n,0);vector<vector<bool>> ispalind(n,vector<bool>(n,false));if(s.size()<=1)return 0;for(i = 0; i < n; ++i){dp[i] = i;ispalind[i][i] = true;if(i < n-1 && s[i]==s[i+1])ispalind[i][i+1] = true;}for(len = 1; len < n; ++len){for(i = 0; i < n-len; ++i){if(ispalind[i][i+len-1] && i-1>=0 && s[i-1]==s[i+len])//是回文串ispalind[i-1][i+len] = true;}}for(i = 1; i < n; ++i){for(j = i; j > 0; --j){if(ispalind[0][i])dp[i] = 0;else if(ispalind[j][i])dp[i] = min(dp[i], dp[j-1]+1);}}return dp[n-1];} };

124 ms 7.4 MB

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