Palindrome Partitioning II

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.

提交好几次都有错误，一调试都是索引边缘条件之类的错误。先dp求出整个字符串的Palindrome情况。然后按长度从1到串长度L来dp求mincut。

关键点在于，假设当前长度为L，那么迭代i(此i为索引值非长度值)从1，2，，，，L-1查看dp(i，L-i)是否Palindrome，如果是那么与mincut比较保存最小值，如果不是，那么直接忽略，可忽略原因是因为在i迭代过程中比当前i大的索引还有机会再计算剩下的dp(i，L-i)是否Palindrome，所以这里可以直接去掉重复计算。

class Solution {public:int minCut(string s) {const int size = s.length();vector< vector<bool> > dp(size, vector<bool>(size+1, 0));for (int len = 1; len <= size; len++){for (int i = 0; i+len <= size; i++){if (len == 1)dp[i][1] = true;else if (len == 2 && s[i] == s[i+1])dp[i][2] = true;else if (len > 2 && dp[i+1][len-2] == 1 && s[i] == s[i+len-1])dp[i][len] = true;}} vector<int> dpmin(size , 0);for (int i = 1; i <= size; i++){int minv = INT_MAX;if (dp[0][i]){dpmin[i-1] = 0;continue;}for (int j = 1; j  < i; j++){if (dp[j][i-j])minv = min(minv, dpmin[j-1] + 1);}dpmin[i-1] = minv;}return dpmin[size - 1];}};

下面是拷的别人的精短代码

class Solution {public:    int minCut(string s) {        int N = s.size();        bool isP[N];        int dp[N];        dp[0] = 0;        for (int i = 1; i < N; ++i)         {            isP[i] = true;            dp[i] = dp[i-1] + 1;            for (int j = 0; j < i; ++j)             {                isP[j] = (s[i] == s[j]) ? isP[j+1] : false; // isP[j] == true   -> [j...i] is a palindrome                                                            // isP[j+1] == true -> [j+1...i-1] is a palindrome                if (isP[j])                    dp[i] = (j == 0) ? 0 : min(dp[i], dp[j-1] + 1);  // dp[i] -> minCount for [0...i]            }        }        return dp[N-1];    }};