Question

假设我有一个只包含数字和特殊字符E的数组。该数组可以包含多个E。只要遇到E，我的函数应该返回当前整个集合中的最小数字。

function OffLineMinimum(strArr) {
    var arr = [],
        len = strArr.length,
        res = [],
        i = 0;
  while (i < len) {
        if (strArr[i] != "E") {
            arr.push(parseInt(strArr[i]));
            i++;
        } else {
        res.push(arr.sort(function(a, b) {
            return a - b;
        })[0]);
        i = i + 2;
        }
    }
    return res.join(",");
}
console.log(OffLineMinimum(["5", "4", "6", "E", "1", "7", "E", "E", "3", "2"]));
// output: 4,1,5

Answer 1

方法首先获取数组中%include "io.mac" .STACK 100H .DATA Numbers DW 3,4,5,2,6,0 msg0 db "Printing values in array",0 msg1 db "Max",0 .CODE .STARTUP PutStr msg0 mov dx, Numbers mov si, dx ;point to array printValues: cmp word [si], 0 je procedure nwln PutInt [si] add si, 2 jmp printValues procedure: push Numbers ;push Number to stack to pass parameter by stack call maxMeth nwln PutStr msg1 nwln PutInt ax nwln PutInt bx nwln complete: .EXIT maxMeth: enter 0,0 ;save old bp and set bp to sp mov si, [bp+4] ;point to array mov ax, [si] ; ax holds max mov bx, [si] ; ax holds max add si,2 ; Increment si to next number ;Now entering loop max: cmp word [si],0 ; checks to see if the number is 0 and if it is, then we are done. je finish ;equals 0, then finished cmp ax, [si] jg testSecondMax ;So if ax is greater than si, then dont assign si to ax and check second max jmp newMax ;otherwise go to new max newMax: mov ax, [si] ; save new max jmp increment ; and increment testSecondMax: cmp bx, [si] jl secondMax jg increment secondMax: mov bx, [si] jmp increment increment: add si, 2 ;now increment si to check the next value and jump back to the main loop. jmp max finish: ;clean up. leave ;pop bp ret ;return的所有索引然后遍历该数组切片并将主数组过滤到该索引，然后得到该子集的min

"E"

Answer 2

根据问题，这似乎是你想要的

var t1 = ["5", "4", "6", "E", "1", "7", "E", "E", "3", "2"];
var t2 = ["5", "4", "6", "E", "8", "7", "E", "E", "3", "2"];
function OffLineMinimum(strArr) {
    return strArr.reduce((result, item) => {
        if (item === 'E') {
            result.output.push(result.work.sort((a, b) => a - b).shift());
        } else {
            result.work.push(Number(item));
        }
        return result;
        
    }, {output: [], work: []}).output;
}
console.log(OffLineMinimum(t1).toString());
console.log(OffLineMinimum(t2).toString());

然而，t2，根据评论应该导致4,7,5

所以逻辑完全没有描述。但似乎连续的第二个E使用到目前为止的剩余完整集，其中第一个应该使用自上一个E以来的当前“块”

代码：

var t1 = ["5", "4", "6", "E", "1", "7", "E", "E", "3", "2"];
var t2 = ["5", "4", "6", "E", "8", "7", "E", "E", "3", "2"];
function OffLineMinimum(strArr) {
    return strArr.reduce((result, item) => {
        if (item === 'E') {
            if (result.lastItem === 'E') {
                result.output.push(result.work.sort((a, b) => a - b).shift());
            } else {
                result.output.push(result.block.sort((a, b) => a - b).shift());
                result.work.push(...result.block);
                result.block = [];
            }
        } else {
            result.block.push(Number(item));
        }
        result.lastItem = item;
        return result;
        
    }, {output: [], work: [], block: [], lastItem: ''}).output;
}
console.log(OffLineMinimum(t1).toString())
console.log(OffLineMinimum(t2).toString())

如何使用JavaScript查找最小的数字？

2 个答案: