Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.
Input: "aba" Output: True
Input: "abca" Output: True Explanation: You could delete the character ‘c‘.
判断字符串是不是回文，最多可以删除字符串的一个元素来使它成为回文。
解决：从字符串两端开始比较，用一个指针low从前遍历，用high从尾部遍历。
1、当s[low] == s[high]，说明符合回文，++low, --high，遍历下一组。
2、要是s[low] != s[high]，就要删除一个数，重组。重组的原则就是s[low]和s[high-1]比较，s[high]和s[low+1]比较，是否能配对。
2.1、要是s[low+1] == s[high]，那就删除当前的s[low]；对应的s[low] == s[high-1]，就删除当前的s[high]。
2.2、要是两个都符合条件，就要顺次看下一对能不能组合。
2.3、要是两个都不符合条件，那就说明只删除一个元素不能形成回文。
class Solution { public: bool validPalindrome(string s) { int low = 0; int high = s.length() - 1; int cnt = 0; while (low<high) { if (s[low] != s[high]) { if (s[low+1] == s[high] && s[low] == s[high-1]) { if (s[low+1] == s[high] && s[low+2] == s[high-1]) ++low; else if (s[low] == s[high-1] && s[low+1] == s[high-2]) --high; } else if (s[low+1] == s[high]) ++low; else if (s[low] == s[high-1]) --high; else return false; ++cnt; } if (cnt==2) return false; ++low; --high; } return true; } };
