Problem 132: Palindrome Partitioning II
思路
这道题的思路就是用一个一维数组去记录最小的cut。
注意去区分和 combination 当中 Palindrome Partitioning 题目的区别。那道题问一共有多少种 cut, 这肯定要用到搜索,排列组合的知识;而这道题问的是最少几种 cut,是最优问题,需要动态规划。
public class Solution {
public int minCut(String s) {
if (s == null || s.length() == 0) {
return 0;
}
boolean[][] isPalindrome = getIsPalindrome(s);
int[] f = new int[s.length() + 1];
f[0] = 0;
for (int i = 1; i <= s.length(); i++) {
f[i] = Integer.MAX_VALUE;
for (int j = 0; j < i; j++) {
if (isPalindrome[j][i - 1]) {
f[i] = Math.min(f[i], f[j] + 1);
}
}
}
return f[s.length()] - 1;
}
private boolean[][] getIsPalindrome(String s) {
boolean[][] isPalindrome = new boolean[s.length()][s.length()];
//initialize
for (int i = 0; i < s.length(); i++) {
isPalindrome[i][i] = true;
}
for (int i = 0; i < s.length() - 1; i++) {
isPalindrome[i][i + 1] = (s.charAt(i) == s.charAt(i + 1));
}
for (int length = 2; length < s.length(); length++) {
for (int start = 0; start + length < s.length(); start++) {
isPalindrome[start][start + length] = isPalindrome[start + 1][start + length - 1] && s.charAt(start) == s.charAt(start + length);
}
}
return isPalindrome;
}
}
易错点
分析下整个代码的流程
初始化f[n]
boolean[][] isPalindrome = getIsPalindrome(s); int[] f = new int[s.length() + 1]; f[0] = 0;
如上图所示,建立f[n],用来截止到目前位置的最小cut
update f[n]
for (int i = 1; i <= s.length(); i++) { f[i] = Integer.MAX_VALUE; for (int j = 0; j < i; j++) { if (isPalindrome[j][i - 1]) { f[i] = Math.min(f[i], f[j] + 1); } } }
f[i]设置成最大值,随后进行更新。首先要保证f[i - 1]是true,才能走过来。
初始化
isPalindrome[][]
//initialize
for (int i = 0; i < s.length(); i++) {
isPalindrome[i][i] = true;
}
for (int i = 0; i < s.length() - 1; i++) {
isPalindrome[i][i + 1] = (s.charAt(i) == s.charAt(i + 1));
}
for (int length = 2; length < s.length(); length++) {
for (int start = 0; start + length < s.length(); start++) {
isPalindrome[start][start + length] = isPalindrome[start + 1][start + length - 1] && s.charAt(start) == s.charAt(start + length);
}
}
首先对角线肯定是true;其次是相邻两个字符必须相等才能为true;最难的就是最后一个更新剩下的矩阵
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