[LeetCode] Palindrome Partitioning II

最新推荐文章于 2019-06-01 11:25:07 发布

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本文介绍了一种利用动态规划算法解决字符串最小回文切分问题的方法。通过预先计算字符串的所有子串是否为回文，并存储这些信息以避免重复计算，进而求得将字符串分割成最少数量的回文子串所需的最小切割数。

Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

保存回文判定结果，动规求解

dp[i]=min(dp[k]+1(s[k+1, i]是回文串, dp[k]+i-k(s[k+1, i]不是回文串)), 0<=k<i.

/*
类似矩阵连乘的动归思路。
dp[i][j]=min(dp[i][k]+dp[k+1][j]+1), i<=k<j.
这个方程的时间是O(n^3)，简化dp[i][j]为dp[i]，表示从0到i的minCut.
dp[i]=min(dp[k]+1(s[k+1, i]是回文串, dp[k]+i-k(s[k+1, i]不是回文串)), 0<=k<i.
*/
class Solution {
public:
    vector<vector<int>> pa;
    bool isPalindrome(string &s, int i, int j) {
        if( i > j) return false;
        if(pa[i][j] != -1) return pa[i][j];
        if(i == j) return pa[i][j] = 1;
        
        if(s[i] != s[j]) return pa[i][j] = 0;
        else {
            if(j-i == 1) return pa[i][j] = 1;
            else return pa[i][j] = isPalindrome(s,i+1,j-1);
        }
    }
    int minCut(string s) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int len = s.length();
        if(len <= 1) return 0;
        
        vector<int> min,vtmp;
        pa.clear();
        for(int i = 0; i < len; i++) {
            min.push_back(0);
            vtmp.push_back(-1);
        }
        for(int i = 0; i < len; i++) pa.push_back(vtmp);
        
        int tmp,ans;
        for(int i = 1; i < len; i++) {
            if(isPalindrome(s,0,i)) {
                min[i] = 0;
            }
            else {
                ans = len+1;
                for(int k = 0; k < i; k++) {
                    if(isPalindrome(s,k+1,i)) tmp = min[k]+1;
                    else tmp = min[k]+i-k;
                    if(tmp < ans) ans = tmp;
                }
                min[i] = ans;
            }
        }
        
        return min[len-1];
        
        
    }
};