leetcode第三周解题总结（5，7，66）

5. Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example:
Input: “babad”
Output: “bab”
Note: “aba” is also a valid answer.
Example:
Input: “cbbd”
Output: “bb”

题意解析：
寻找字符串中的最长回文子串

解题思路：
第一周做过一个类似的判断是否为回文字串的题目(125. Valid Palindrome)，但这道题似乎就没那么简单，要找到最长的回文字串。一个思路是遍历所有的中心对称点，如果有n个字符，那么就有(2n - 1)个中心对称点，在每个对称点上用两个指针的方法寻找最长的回文字串，最后输出最长的字符串。

class Solution {public:    string longestPalindrome(string s) {        size_t len = s.size();        if(len == 0 || len == 1) return s;         int left = 0, right = 0;        for (size_t point = 0; point < len; point++) {            int start = point;            int end = point;            while(start >= 0 && end < len && s[start]==s[end]){                start --;                end ++;            }            if(end - start > right - left) {                left = start;                right = end;            }            start = point;            end = point + 1;            while(start >= 0 && end < len && s[start]==s[end]){                start --;                end ++;            }            if(end - start > right - left) {                left = start;                right = end;            }        }        return s.substr(left + 1, right - 1 -left);    }};

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开始第二个知识点的学习 — Math

7. Reverse Integer

Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

题意解析：
32位有符号整数的翻转，如果溢出则返回0

解题思路：
数学题，思路并不难，直接看代码即可。

class Solution {public:    int reverse(int x) {        int y = 0;        while (abs(x) > 0) {            if(abs(y) > INT_MAX/10 || (abs(y) == INT_MAX/10  && (x >= 7 || x <= -8))) return 0;            y = y*10 + x%10;            x = x/10;        }        return y;    }};

66. Plus One

Given a non-negative integer represented as a non-empty array of digits, plus one to the integer.
You may assume the integer do not contain any leading zero, except the number 0 itself.
The digits are stored such that the most significant digit is at the head of the list.

题意解析：
非负整数加1，进位操作

解题思路：
只有当该数字每一位上都是9才会增加一位，而其他情况则比较简单，直接进位操作

class Solution {public:    vector<int> plusOne(vector<int>& digits) {        size_t len =  digits.size();        for (int i = len - 1; i >= 0; i--) {            if (digits[i] < 9) {                digits[i] += 1;                return digits;            } else {                digits[i] = 0;            }        }        digits[0] = 1;        digits.push_back(0);        return digits;    }};