Leetcode算法学习日志-680 valid Palindrome II
2017-09-18 16:16
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Leetcode-680 Valid Palindrome II
题目原文
Given a non-empty strings, you may delete
at most one character. Judge whether you can make it a palindrome.
Example 1:
Input: "aba" Output: True
Example 2:
Input: "abca" Output: True Explanation: You could delete the character 'c'.
Note:
The string will only contain lowercase characters a-z.The maximum length of the string is 50000.
题意分析
判断一个字符串在最多删去一个字符的情况下是否能成为回文,找最长回文的算法复杂度为O(n^2),而这道题只需要判断是不是回文,复杂度应该控制在O(n)。解法分析
如果遍历每一个可以删除的元素,再判断剩下字符串是否为回文,算法复杂度为O(n^2),注意对于string的==运算复杂度为o(n)。应该用两个指针i,j分别从string两端往中间遍历,如果相等就继续,如果不能,记录现有i,j,跳过其中一个字符,继续判断。判断过程中需要保存几个值way、count,用于记录回退的次数,比如跳过某个字符后继续进行遍历,此时way应变为1,如果再遇到不想等元素,则回退,way=2,如果再一次遇到不相等元素,则返回false,如果循环正常退出,返回true。C++代码如下:class Solution {
public:
bool validPalindrome(string s) {
int i,j;
int si,sj;
int n=s.size();
int count=0;//if count reaches 1, there is no chance to delete a character. But you can change the way
int way=0;//if way reaches 2 and count is 1, when you find a different char, you return false
for(i=0,j=n-1;i<j;i++,j--){
if(s[i]==s[j])
continue;
else{
if(way==2){
return false;
}
if(way==1){
way++;
i=si;
j=sj;
j++;
}
if(way==0){
way++;
count=1;
si=i;
sj=j;
i--;
}
}
}
return true;
}
};
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