132 Palindrome Partitioning II

132 Palindrome Partitioning II

这道题就是标识出s[i:j+1]是否为palindrome， 然后dp找出最小分割

class Solution:
    # @param {string} s
    # @return {integer}
    def minCut(self, s):
        LS = len(s)
        dp = [[False] * LS for i in range(0, LS)]
        ans = [1<<31 for i in range(0 ,LS)]
        for ls in range(0, LS):
            for i in range(0, LS-ls):
                if s[i] == s[i+ls] and (ls <= 2 or dp[i+1][i+ls-1]):
                    dp[i][i+ls] = True
        ans[0] = 0
        for i in range(1, LS):
            for j in range(0,i+1):
                if dp[j][i]:
                    ans[i] = 0 if j == 0 else (min(ans[i], 1+ans[j-1]))
        return ans[-1]