TypeScript: 判断两个数组的内容是否相等

一、直接遍历

// 1.直接遍历
const arr1: any[] = ["apple", "banana", NaN];
const arr2: any[] = ["apple", NaN, "banana"];function fn1(arr1: any[], arr2: any[]) {// Array.some(): 有一项不满足，返回falseif (arr1.length !== arr2.length) {return false;}return arr1.some((item: any) => arr2.indexOf(item) === -1) ? false : true;
//   return !arr1.some((item) => !arr2.includes(item));
}
console.log("直接遍历 => ", fn1(arr1, arr2)); // false

NaN 会有问题

优化版本

// 1.直接遍历
const arr1: any[] = ["apple", "banana", NaN];
const arr2: any[] = ["apple", NaN, "banana"];function fn1(arr1: any[], arr2: any[]) {// Array.some(): 有一项不满足，返回falseif (arr1.length !== arr2.length) {return false;}
//   return arr1.some((item: any) => arr2.indexOf(item) === -1) ? false : true;return !arr1.some((item) => !arr2.includes(item));
}
console.log("直接遍历 => ", fn1(arr1, arr2)); // true

Array.prototype.indexOf() 是使用的严格相等算法 => NaN值永远不相等
Array.prototype.includes() 是使用的零值相等算法 => NaN值视作相等

二、把重复元素编号

// 2.把重复元素标识编号
function fn3(arr1: any[], arr2: any[]) {if (arr1.length !== arr2.length) {return false;}// 重复数组元素 加1、2、3const countArr1 = updateArray(arr1);const countArr2 = updateArray(arr2);// arr 数组 元素重复转换成 val1，val2function updateArray(arr: any[]) {const countMap: Map<any, number> = new Map();const updateArr: any[] = [];for (const ele of arr) {//   const cnt = !countMap.has(ele) ? 1 : countMap.get(ele)! + 1;const cnt = (countMap.get(ele) || 0) + 1;countMap.set(ele, cnt);updateArr.push(`${ele}${cnt}`); // 有问题// 等同于//   if (!countMap.has(ele)) {//     // 如果元素是第一次出现，直接添加到结果数组//     countMap.set(ele, 0)//     updateArr.push(ele)//   } else {//     // 如果元素已经出现过，添加带有编号的新元素到结果数组//     const count = countMap.get(ele)! + 1//     countMap.set(ele, count)//     updateArr.push(`${ele}${count}`)//   }}return updateArr;}// console.log("countArr1 => ", countArr1);// console.log("countArr2 => ", countArr2)return !countArr1.some((item) => !countArr2.includes(item));
}const array1 = ["apple", "banana", "cherry", "banana"];
const array2 = ["banana", "apple", "banana", "cherry"];
console.log("把重复元素标识编号 => ", fn3(array1, array2)); // true// 其实这种存在漏洞的
const array3 = ["apple", "banana", "cherry", "banana", 1, "1", "1"];
const array4 = ["banana", "apple", "banana", "cherry", "1", 1, 1];
// 应该是false
console.log("把重复元素标识编号 存在漏洞 => ", fn3(array3, array4)); // true

三、统计元素次数

// 3.统计元素次数
function fn4(arr1: any[], arr2: any[]) {if (arr1.length !== arr2.length) {return false;}// 创建计数对象，用于记录每个元素在数组中的出现次数const countMap1: Map<any, number> = count(arr1);const countMap2: Map<any, number> = count(arr2);function count(arr: any[] = []) {const map: Map<any, number> = new Map();for (const item of arr) {map.set(item, (map.get(item) || 0) + 1);}return map;}//检查计数对象是否相等//   console.log("countMap1 => ", countMap1)
//   console.log("countMap2 => ", countMap2)for (const [key, count] of countMap1) {if (countMap2.get(key) !== count) {return false;}}return true;
}const array5 = ["apple", "banana", "cherry", "banana", 1, "1", "11", 11];
const array6 = ["banana", "apple", "banana", "cherry", "1", 1, "11", 11];console.log("统计元素次数 => ", fn4(array5, array6)); // true
console.log("统计元素次数 => ", fn4(array3, array4)); // false

四、+1，-1

// 4.+1,-1
function fn5(arr1: any[], arr2: any[]) {if (arr1.length !== arr2.length) {return false;}const countMap: Map<any, number> = new Map();// 计算第一个数组的元素for (const item of arr1) {countMap.set(item, (countMap.get(item) || 0) + 1);}// 比较第二个数组for (const item of arr2) {const val = countMap.get(item)if (val === undefined || val <= 0) {return false;}countMap.set(item, val - 1)}return true;
}
console.log("+1, -1 => ", fn5(array3, array4)) // false
console.log("+1, -1 => ", fn5(array5, array6))  // true