leetcode Problem 5 最长子回文串

题目：Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

1、暴力做法。
遍历所有的子串，求出最长。复杂度O(n3)。显然GG了。
2、动态规划
把一个大问题分解成每一个小问题进行求解。假设字符串s，则s[i]到s[j]是回文的话，s[i+1]到s[j-1]也是回文；或者说，已知s[i+1]到s[j-1]是回文的情况下，如果s[i]==s[j]则，s[i]到s[j]就是回文，我们用数组table[i][j]来表示i到j的是否回文状态。代码如下，复杂度是o(n2)。

var longestPalindrome = function (s) {    const length = s.length;    let table = [];    for (let i = 0; i < 1000; i++) {        table[i] = new Array();        for (let j = 0; j < 1000; j++) {            table[i][j] = 0;        }    }    let maxlength = 1,        maxstr,        longestbegin;    for (let i = 0; i < length; i++) {        table[i][i] = 1; //所有的 i 到 i的字符串是回文        longestbegin = i;            }    for (let i = 0; i < length - 1; i++) {        //长度为2的回文        if (s[i] === s[i + 1]){            table[i][i + 1] = 1;            maxlength = 2;            longestbegin = i;        }    }    //状态转移 再table[i+1][j-1] == 1的情况下，如果i === j，则 table[i][j]为1 从长度为3的回文开始    for (let len = 3; len <= length; len++) {        for (let i = 0; i < length - len + 1; i++) {            //遍历所有长度为len的字符串            let j = i + len - 1;//两种情况 单数回文和双数回文            //let j = len % 2 === 0?i + len - 2            if (s[i] === s[j] && table[i + 1][j - 1] === 1) {                table[i][j] = 1;                maxlength = len;                longestbegin = i;            }        }    }    //console.log(longestbegin, maxlength)    return s.slice(longestbegin,longestbegin +  maxlength)};

然而，这里还是过不了leetcode的。显然，还需要优化的地方
好像o(n2)也是能强行过的。。把字符串换成数组再处理。不过时间只优于百分之1的人，太菜了23333

//最长回文串 o(n2)var longestPalindrome = function (s) {    const length = s.length;    s = s.split('');    let table = [];    let maxlength = 1,        maxstr,        longestbegin = 0;    for (let i = 0; i < length; i++) {        table[i] = new Array();        table[i][i] = 1;    }    for (let i = 0; i < length - 1; i++) {        //长度为2的回文        if (s[i] === s[i + 1]) {            table[i][i + 1] = 1;            maxlength = 2;            longestbegin = i;        }    }    //状态转移 再table[i+1][j-1] == 1的情况下，如果i === j，则 table[i][j]为1 从长度为3的回文开始    for (let len = 2; len <= length; len++) {        for (let i = 0; i < length - len + 1; i++) {            //遍历所有长度为len的字符串            let j = i + len - 1; //两种情况 单数回文和双数回文            //let j = len % 2 === 0?i + len - 2            if (s[i] === s[j] && table[i + 1][j - 1] === 1) {                table[i][j] = 1;                maxlength = len;                longestbegin = i;            }        }    }    //console.log(longestbegin, maxlength)    return s.join('').slice(longestbegin, longestbegin + maxlength)//真的是日了狗。。。数组处理比字符串处理要快};

而且不是每次都能过，这个还是要优化的啊

换另一种思路，就是。遍历字符串中的每个字符。算出以该字符为中心的最长回文串

var longestPalindrome2 = function (s) {    function expandAroundCenter(s, l, r) {        while (l >= 0 && r < sLen && s[l] == s[r]) {            l--;            r++;        }        return r - l - 1;    }    let start = 0,        end = 0,        len1 = 0,        len2 = 0,        len = 0,        sLen = s.length;    for (let i = 0; i < sLen; i++) {        len1 = expandAroundCenter(s, i, i);//aba这种回文        len2 = expandAroundCenter(s, i, i + 1);//abba这种回文        len = Math.max(len1, len2);//算出较大的回文        if (len > end - start) {            end = i + (len >> 1);            start = end + 1 - len;        }    }    return s.substring(start, end + 1);};

这次就快多了。。。。