public class Solution {
public boolean validPalindrome(String s) {
//如果字符串为空或者长度为1，直接返回true
if ("".equals(s) || 1 == s.length()) {
return false;
}
//初始化指针
int left = 0;
int right = s.length() - 1;
//初始化删除标记位
int flag = 0;
while (left < right) {
//对称位置相等，循环继续
if (s.charAt(left) == s.charAt(right)) {
left++;
right--;
continue;
}
//对称位置不等，判断标记位，为0代表尚未使用过删除功能，则继续
if (flag == 0) {
//如果左右只差一位，则删除任意一个皆可，返回true
if (left + 1 == right) {
return true;
}
//这里需要多看一步，否则会将如"bcaacab"判断为false，错误示例如下
//step1: b == b , continue;
//step2: c != a , flag == 0 , a == a, flag ++, continue;
//step3：a != c , flag == 1 , return false.
if (s.charAt(left + 1) == s.charAt(right) && s.charAt(left + 2) == s.charAt(right - 1)) {
left += 2;
right--;
flag++;
continue;
}
//此处与上处类似，不再赘述
if (s.charAt(left) == s.charAt(right - 1) && s.charAt(left + 1) == s.charAt(right - 2)) {
left++;
right -= 2;
flag++;
continue;
}
}
return false;
}
return true;
}
}
