190927+191001 #20
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190927+191001 #20
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Flynnon
reviewed
Oct 11, 2019
| var searchRange = function(nums, target) { | ||
| let num = [-1,-1] | ||
| let a = 0 | ||
| if(nums.indexOf(target) === -1) return num |
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这一步没必要，indexOf 实际上也是遍历查找，严重影响性能。
| last = null, | ||
| flag1 = true | ||
| flag2 = true | ||
| for(var i = 0,j = nums.length-1;i <nums.length,j >= 0;i++,j--) { |
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i <nums.length,j >= 0
这个是什么意思? js有这种语法么?
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如果这种方法的话，这里少了一个退出条件， i <= j。可以避免一些无效比较。
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| flag1 = true | ||
| flag2 = true |
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仅个人感觉..这里的flag是不是初始值用 false，找到后更新为 true 好一些..
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用true是因为false就可以跳出来了
WangXinShu
reviewed
Oct 14, 2019
leetcode/bsy/简单/14-190929-最长公共前缀.js
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| var longestCommonPrefix = function(strs) { | ||
| let i = 1 | ||
| let item = '' | ||
| if(strs.length>0){ |
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strs.length === 0 可以直接返回
WangXinShu
reviewed
Oct 14, 2019
leetcode/bsy/简单/14-190929-最长公共前缀.js
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| let i = 1 | ||
| let item = '' | ||
| if(strs.length>0){ | ||
| let len = strs[0].split('').length |
WangXinShu
reviewed
Oct 14, 2019
leetcode/bsy/简单/14-190929-最长公共前缀.js
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| if(strs.length === 1){ | ||
| return strs[0] | ||
| }else if(len === 1){ | ||
| for(let j = 1; j < strlen; j++){ | ||
| if (strs[j].search(strs[0]) !== 0) { | ||
| return item = '' | ||
| }else { | ||
| item = strs[0] | ||
| } | ||
| } | ||
| }else { | ||
| for(i;i<=len;i++){ | ||
| let main = strs[0].substr(0,i) | ||
| for(let j = 1; j < strlen; j++){ | ||
| if (strs[j].search(main) !== 0) { | ||
| return item.substr(0,i-1) | ||
| }else { | ||
| item = main | ||
| } | ||
| } | ||
| } | ||
| } | ||
| } |
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代码写的太麻烦了，就从第一个元素的第一个字母开始，遍历数组，判断是否所有的元素的第一个元素都一样，再取第二个。。。。
WangXinShu
reviewed
Oct 14, 2019
leetcode/bsy/简单/14-190929-最长公共前缀.js
Outdated
| let i = 1 | ||
| let item = '' | ||
| if(strs.length>0){ | ||
| let len = strs[0].split('').length |
WangXinShu
approved these changes
Oct 14, 2019
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