摘要：工具安裝使用示例問題下面兩部分的順序不能交換第一個(gè)參數(shù)是空數(shù)組故意寫錯(cuò)答案，展示測(cè)試失敗輸出效果測(cè)試用例編寫說明要測(cè)試的都是函數(shù)，參數(shù)個(gè)數(shù)不定，但返回值是一個(gè)。上面例子的輸入?yún)?shù)是，第一個(gè)參數(shù)是數(shù)組，第二個(gè)參數(shù)是數(shù)值返回值是一個(gè)數(shù)組。

最近在用es6解leetcode，當(dāng)問題比較復(fù)雜時(shí)，有可能修正了新的錯(cuò)誤，卻影響了前面的流程。要用通用的測(cè)試工具，卻又有殺雞用牛刀的感覺，所以就寫了個(gè)簡(jiǎn)單易用的leetcode開發(fā)測(cè)試工具，分享與大家。

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] === "*") {
        //下面兩部分的順序不能交換
        return firstMath && isMatch(s.substring(1), p) || isMatch(s, p.substring(2))
    } else {
        return firstMath && isMatch(s.substring(1), p.substring(1))
    }
};
let cases = [              // [[[],""],],   //第一個(gè)參數(shù)是空數(shù)組
    [["abbabaaaaaaacaa", "a*.*b.a.*c*b*a*c*"], true],
    [["aaa", "a*ac"], true],                //故意寫錯(cuò)答案，展示測(cè)試失敗輸出效果
    [["a", "..*"], true],
]
test(isMatch, cases)

測(cè)試用例編寫說明

leetcode要測(cè)試的都是函數(shù)，參數(shù)個(gè)數(shù)不定，但返回值是一個(gè)。因此，我設(shè)計(jì)用例的輸入形式為一個(gè)用例就是一個(gè)兩個(gè)元素的數(shù)組，第一個(gè)元素是一個(gè)數(shù)組：對(duì)應(yīng)輸入?yún)?shù)；第二個(gè)元素是一個(gè)值。
上面例子的輸入?yún)?shù)是（[2, 7, 11, 15], 91），第一個(gè)參數(shù)是數(shù)組，第二個(gè)參數(shù)是數(shù)值；返回值是一個(gè)數(shù)組（[0, 1]）。 如果要測(cè)試的函數(shù)的輸入?yún)?shù)就是一個(gè)數(shù)組，要注意輸入形式，比如，求[1,2,3,4]平均值，要這樣輸入測(cè)試用例： [[[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;                    //從隊(duì)列頭部開始
        let three = nums.length - 1;        //從隊(duì)列尾部開始

        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 = [               // [[[],""],],   //第一個(gè)參數(shù)是空數(shù)組
    [[[]],[]],
    [[[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)

測(cè)試用例編寫說明

測(cè)試用例的7與8，期待結(jié)果的數(shù)組元素順序并不影響答案的判定。

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，已經(jīng)把輸出的樣子調(diào)得還能看過眼了，答案對(duì)比算法，也改進(jìn)了。遇到問題，我會(huì)持續(xù)改進(jìn)，大家遇到問題也可提bug給我，我會(huì)盡快處理。