LintCode 108.[Python] 分割回文串 II 给定字符串 s, 需要将它分割成一些子串, 使得每个子串都是回文串.最少需要分割几次?

LintCode 108. 分割回文串 II

给定字符串 s, 需要将它分割成一些子串, 使得每个子串都是回文串.最少需要分割几次?

Code

def minCut(self, s):
    # write your code here
    if len(s)==0:
        return 0
    n=len(s)
    Judge=[[False for _ in range(n)] for _ in range(n)]
    for i in range(0,n,1):
        for j in range(i,n,1):
            Judge[i][j]=False
    for j in range(n):#奇数
        i = j
        while i >= 0 and j < n and s[i] == s[j]:
            Judge[i][j] = True
            i -= 1
            j += 1
    for j in range(1, n): #偶数
        i = j - 1
        while i >= 0 and j < n and s[i] == s[j]:
            Judge[i][j] = True
            i -= 1
            j += 1
    dp=[[0] for _ in range(n+1)]
    dp[0]=0
    for j in range(1, n+1,1):
        dp[j] = float('inf')
        for i in range(0,j,1):
            if Judge[i][j-1]:#是不是回文字符串
                dp[j] = min(dp[j], dp[i] + 1)
        
    
    return dp[n]-1#要求，最少需要分割几次?

想法

1.dp[i] + 1 为前i个字符可以划分成几个字符串。
2.最后一步就是最后一段