Write a function to find the longest common prefix string amongst an array of strings. If there is no common prefix, return an empty string`""`. **Example 1:** ~~~ Input: ["flower","flow","flight"] Output: "fl" ~~~ **Example 2:** ~~~ Input: ["dog","racecar","car"] Output: "" Explanation: There is no common prefix among the input strings. ~~~ **Note:** All given inputs are in lowercase letters`a-z`. ``` var longestCommonPrefix = function(strs) { if(!strs.length) return ''; var len = strs.reduce((pre,item)=>{ return Math.min(pre,item.length); },Number.MAX_VALUE); var ans=''; for(var i=0;i<len;i++){ var a = strs[0][i]; var f = strs.every((item)=>{ return item[i] === a; }); if(!f) break; ans += a; } return ans; }; ``` ``` var longestCommonPrefix = function(strs) { if (!strs.length) return "" let baseWord = strs[0]; for (let word = 1; word < strs.length; word++) { // currWord = String(strs[word]) while (String(strs[word]).indexOf(baseWord) !== 0 && baseWord !== "") { baseWord = baseWord.substring(0, baseWord.length - 1) } } return baseWord; }; ``` ``` var longestCommonPrefix = function(strs) { if (strs.length == 0) return ""; let prefix = strs[0]; for (let i = 1; i < strs.length; i++) while (strs[i].indexOf(prefix) != 0) { prefix = prefix.slice(0, prefix.length - 1); if (prefix==="") return ""; } return prefix; }; ``` reduce() 方法接收一个函数作为累加器(accumulator)，数组中的每个值(从左到右)开始缩减，最终为一个值。 reduce() 为数组中的每一个元素依次执行回调函数，不包括数组中被删除或从未被赋值的元素，接受四个参数：初始值(或者上次回调函数的返回值)，当前元素值，当前索引，调用reduce的数组。 ``` arr.reduce(callback, [initialValue]) ``` callback (执行数组中每个值的函数，包含四个参数) - previousValue (上一次调用回调返回的值，或者是提供的初始值(initialValue)) - currentValue (数组中当前被处理的元素) - index (当前元素在数组中的索引) - array (调用 reduce 的数组) initialValue (作为第一次调用 callback 的第一个参数) ``` var arr = [1, 2, 3, 4]; var sum = arr.reduce(function(pre, cur, index, arr) { console.log(pre, cur, index); return pre + cur; }) console.log(arr, sum); ```