aduocd 2019-06-29
最近在用es6解leetcode,当问题比较复杂时,有可能修正了新的错误,却影响了前面的流程。要用通用的测试工具,却又有杀鸡用牛刀的感觉,所以就写了个简单易用的leetcode开发测试工具,分享与大家。
npm i leetcode_test
codes:
let test = require('leetcode_test').test /** * @param {string} s * @param {string} p * @return {boolean} */ var isMatch = function (s, p) { if (p.length === 0) { return s.length === 0 } firstMath = s.length > 0 && (p[0] === s[0] || p[0] === '.') if (p.length >= 2 && p[1] === '*') { //下面两部分的顺序不能交换 return firstMath && isMatch(s.substring(1), p) || isMatch(s, p.substring(2)) } else { return firstMath && isMatch(s.substring(1), p.substring(1)) } }; let cases = [ // [[[],''],], //第一个参数是空数组 [['abbabaaaaaaacaa', 'a*.*b.a.*c*b*a*c*'], true], [['aaa', 'a*ac'], true], //故意写错答案,展示测试失败输出效果 [['a', '..*'], true], ] test(isMatch, cases)测试用例编写说明
leetcode要测试的都是函数,参数个数不定,但返回值是一个。因此,我设计用例的输入形式为一个用例就是一个两个元素的数组,第一个元素是一个数组:对应输入参数;第二个元素是一个值。
上面例子的输入参数是([2, 7, 11, 15], 91),第一个参数是数组,第二个参数是数值;返回值是一个数组([0, 1])。 如果要测试的函数的输入参数就是一个数组,要注意输入形式,比如,求[1,2,3,4]平均值,要这样输入测试用例: [[[1,2,3,4]],2.5]
out:
test [1] success, Input: ('abbabaaaaaaacaa','a*.*b.a.*c*b*a*c*'); Expected: true; Output: true test [2] fail, Input: ('aaa','a*ac'); Expected: true; Output: false test [3] success, Input: ('a','..*'); Expected: true; Output: true Result: test 3 cases, success: 2, fail: 1 running 5 ms
codes:
let test = require('leetcode_test').test /** * @param {number[]} nums * @return {number[][]} */ var threeSum = function (nums) { nums = nums.sort((a,b) => a - b); const rs = []; let i = 0; while (i < nums.length) { let one = nums[i]; let two = i + 1; //从队列头部开始 let three = nums.length - 1; //从队列尾部开始 while (two < three) { let sum = one + nums[two] + nums[three]; if (sum === 0) { rs.push([one,nums[two],nums[three]]); two++; three--; while (two < three && nums[two] === nums[two - 1]) { two++; } while (two < three && nums[three] === nums[three + 1]) { three--; } } else if (sum > 0) three--; else two++; } i++; while (i < nums.length && nums[i] === nums[i - 1]) i++; } return rs; }; let cases = [ // [[[],''],], //第一个参数是空数组 [[[]],[]], [[[1,-1,-1,0]],[-1,0,1]], [[[-1,0,1,0]],[[-1,0,1]]], [[[0,0,0,0]],[0,0,0]], [[[-1,2,-1]],[-1,-1,2]], [[[0,0,0]],[0,0,0]], [[[-1,0,1,2,-1,-4]],[[-1,-1,2],[-1,0,1]]], //answer's sequence is not important [[[-1,0,1,2,-1,-4]],[[-1,0,1],[-1,-1,2]]], //answer's sequence is not important [[[-4,-2,-2,-2,0,1,2,2,2,3,3,4,4,6,6]],[[-4,-2,6],[-4,0,4],[-4,1,3],[-4,2,2],[-2,-2,4],[-2,0,2]]], [[[-4,-2,1,-5,-4,-4,4,-2,0,4,0,-2,3,1,-5,0]],[[-5,1,4],[-4,0,4],[-4,1,3],[-2,-2,4],[-2,1,1],[0,0,0]]], ] test(threeSum,cases)测试用例编写说明
测试用例的7与8,期待结果的数组元素顺序并不影响答案的判定。
out:
test [1] success, Input: ([]); Expected: []; Output: [] test [2] success, Input: ([-1,-1,0,1]); Expected: [-1,0,1]; Output: [[-1,0,1]] test [3] success, Input: ([-1,0,0,1]); Expected: [[-1,0,1]]; Output: [[-1,0,1]] test [4] success, Input: ([0,0,0,0]); Expected: [0,0,0]; Output: [[0,0,0]] test [5] success, Input: ([-1,-1,2]); Expected: [-1,-1,2]; Output: [[-1,-1,2]] test [6] success, Input: ([0,0,0]); Expected: [0,0,0]; Output: [[0,0,0]] test [7] success, Input: ([-4,-1,-1,0,1,2]); Expected: [[-1,-1,2],[-1,0,1]]; Output: [[-1,-1,2],[-1,0,1]] test [8] success, Input: ([-4,-1,-1,0,1,2]); Expected: [[-1,-1,2],[-1,0,1]]; Output: [[-1,-1,2],[-1,0,1]] test [9] success, Input: ([-4,-2,-2,-2,0,1,2,2,2,3,3,4,4,6,6]); Expected: [[-2,-2,4],[-2,0,2],[-4,-2,6],[-4,0,4],[-4,1,3],[-4,2,2]]; Output: [[-2,-2,4],[-2,0,2],[-4,-2,6],[-4,0,4],[-4,1,3],[-4,2,2]] test [10] success, Input: ([-5,-5,-4,-4,-4,-2,-2,-2,0,0,0,1,1,3,4,4]); Expected: [[-2,-2,4],[-2,1,1],[-4,0,4],[-4,1,3],[-5,1,4],[0,0,0]]; Output: [[-2,-2,4],[-2,1,1],[-4,0,4],[-4,1,3],[-5,1,4],[0,0,0]] Result: test 10 cases, success: 10, fail: 0
工具地址:https://github.com/zhoutk/lee...
解答地址:https://github.com/zhoutk/lee...
最近一直在用,已经把输出的样子调得还能看过眼了,答案对比算法,也改进了。遇到问题,我会持续改进,大家遇到问题也可提bug给我,我会尽快处理。