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.
f(i,j) 表示[i,j]之前最小的cut数，状态转移方程：
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.
f(i,j) 表示[i,j]之前最小的cut数，状态转移方程：
f(i,j) = min{f(i,k) + f(k+1,j)} i<=k<=j, 0<=i <=j <n
这是一个二维函数，所以我们转化为一维DP
f(i) 表示[i, n-1] 之间最小的cut数， n为字符串长度，则状态转移方程为
f(i) = min {f(j+1) +1} , i<=j < n
判断[i,j]是否是回文
P[i,j] = true if [i,j]是回文， 那么
P[i][j] = str[i] == str[j] && P[i+1][j-1]
class Solution {
public:
int minCut(string s) {
const int n = s.size();
int f[n+1];
bool p[n][n];
fill_n(&p[0][0], n*n, false);
for (int i = 0; i <=n; i++) {
f[i] = n-1-i;
}
for (int i = n-1; i >=0; i--) {
for (int j = i; j < n; j++) {
if (s[i] == s[j] && (j-i < 2 || p[i+1][j-1])) {
p[i][j] = true;
f[i] = min(f[i], f[j+1]+1);
}
}
}
return f[0];
}
};版权声明：本文为博主原创文章，未经博主允许不得转载。