Question

我正在努力解决一个棘手的问题。每当有3行（例如3A，3B和3C）时，我想增加变量'count'。我的逻辑仅适用于第一行，但其余部分则失败。这个问题的答案应该是3.有人可以在这里说清楚吗？

function doThis() {
    var everySeat = [1A, 1B, 1C, 2A, 2C, 3A, 3B, 3C, 151A, 151B, 151C, 152B, 152C];
    var count;

    for (i = 1; i <= everySeat.length; i++) {
        if (everySeat[i-1] = i + ‘A’ && everySeat[i] = i + ‘B’ && everySeat[i+1] = i + ‘C’) {
            count++;
        }
    }

}

doThis();

Answer 1

试试这个。工作正常。

function doThis() {
    var everySeat = ['1A', '1B', '1C', '2A', '2C', '3A', '3B', '3C', '151A', '151B', '151C', '152B', '152C'];
    var count = 0;

    for (var i = 1; i <= everySeat.length-2; i++) {
        var prefix = parseInt(everySeat[i-1]);
        if (everySeat[i-1] == prefix + 'A' && everySeat[i] == prefix + 'B' && everySeat[i+1] == prefix + 'C') {
            count++;
        }
    }
return count;
}

Answer 2

首先，您的数组有语法错误。所以尝试下面的代码。

var everySeat = ['1A', '1B', '1C', '2A', '2C', '3A', '3B', '3C', '151A', '151B', '151C', '152B', '152C'];
    var count =0;
    for(var i =0; i < everySeat.length ; i++ ){
    if(everySeat[i] == i + 'A' && everySeat[i+1] == i + 'B' && everySeat[i+2] == i + 'C'  ){
    count++;
    }
    }
    console.log(count);

Answer 3

具有Array.slice()和String.match()功能的扩展解决方案：

var everySeat = ['1A', '1B', '1C', '2A', '2C', '3A', '3B', '3C', '151A', '151B', '151C', '152B', '152C'],
    count = 0;
	
for (var i=0, l=everySeat.length; i < l-2; i++) {
    if (String(everySeat.slice(i,i+3)).match(/\d+A,\d+B,\d+C/)) {
        count++;
        i +=2;
    }	
}

console.log(count);

Answer 4

您可以将其转换为字符串，然后匹配所需的子字符串出现。

var everySeat = ['1A', '1B', '1C', '2A', '2C', '3A', '3B', '3C', '151A', '151B', '151C', '152B', '152C'];
var str = everySeat.toString();
var count1 = (str.match(/3A,3B,3C/g) || []).length;
console.log(count1);

Answer 5

您可以使用哈希表来计算行数，如果行数为3，则可以添加一个哈希表。

此提案也适用于未分类的数据。

var array = ['1A', '1B', '1C', '2A', '2C', '3A', '3B', '3C', '151A', '151B', '151C', '152B', '152C'],
    count = Object.create(null),
    result = array.reduce(function (r, s) {
        var key = s.match(/\d+/)[0];
        count[key] = (count[key] || 0) + 1;
        return r + (count[key] === 3);
    }, 0);

console.log(result);

从数组中选择特定元素

5 个答案: