获取数组中的最大值(最小值)

如题所示，开始敲代码

方案一：给数组先排序(由大到小排序),第一项就是最大值，最后一项是最小值

let ary = [12, 13, 14, 23, 24, 13, 15, 12];
let max = ary.sort((a, b) => b - a)[0];
let min = ary.sort((a, b) => b - a)[ary.length - 1];
console.log(max, min);   // 24 12

方案二：假设第一个值是最大值/最小值,依次遍历数组中后面的每一项,和假设的值进行比较,如果比假设的值要大/小,把当前项赋值给MAX/MIN

let ary = [12, 13, 14, 23, 24, 13, 15, 12];
let max = ary[0],
    min = ary[0];
for (let i = 1; i < ary.length; i++) {
    let item = ary[i];
    item > max ? max = item : null;
    item < min ? min = item : null;
}
console.log(max, min);   // 24 12

方案三：基于Math.max/Math.min配合apply完成

let ary = [12, 13, 14, 23, 24, 13, 15, 12];
let max = Math.max.apply(null,ary);
let min = Math.min.apply(null,ary)
console.log(max, min);   // 24 12

方案四：基于Math.max/Math.min配合eval完成

let ary = [12, 13, 14, 23, 24, 13, 15, 12];
let max = eval("Math.max("+ary.join()+")");
let min = eval("Math.min("+ary.join()+")");
console.log(max, min);   // 24 12

方案五：基于Math.max/Math.min配合ES6展开运算符完成

let ary = [12, 13, 14, 23, 24, 13, 15, 12];
let max = Math.max(...ary);
let min = Math.min(...ary)
console.log(max, min);   // 24 12