Question

我有数组

array = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]];

我希望找到子数组的索引

[1, 2, 3, 4]

所以它应该返回值0
但每次我尝试这样做，它返回-1

这是我试过的：

array.indexOf(1, 2, 3, 4);

我知道这是因为我正在搜索的子阵列中有逗号我该怎么办？

Answer 1

这是一个快速而肮脏的解决方案，它将数组作为字符串进行比较：

var array = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]];

var target = [9, 10, 11, 12].toString();
for(var i=0; i<array.length; i++) {
   if(target === array[i].toString()) break;
}

if(i >= array.length) {
    console.log("not found");
} else {
    console.log("found at index " + i);
}

http://jsfiddle.net/jMfwQ/

Answer 2

问题在于，当javascript使用==来比较数组时，它无法按预期工作。您必须构建自己的数组相等检查器（请参阅How to check if two arrays are equal with JavaScript?）。然后遍历数组并检查每个子元素是否等于您要查找的元素

function arraysEqual(a, b) {
    if (a === b) return true;
    if (a == null || b == null) return false;
    if (a.length != b.length) return false;

    // If you don't care about the order of the elements inside
    // the array, you should sort both arrays here.

    for (var i = 0; i < a.length; ++i) {
        if (a[i] !== b[i]) return false;
    }
    return true;
}

var haystack = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]];
var needle = [1, 2, 3, 4];
var i = 0;

for(var sub in haystack) {
    if(array_equal(sub, needle)) { // found it
        return i;
    }
    i++;
}

if(i >= haystack.size) //didn't find it
    i = -1;

return i; //return the index of the found needle-array

Answer 3

一个死简单的字符串匹配实现，它只使用本机方法而不使用变量：

[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
 .map( /./.test, RegExp( "^"+ [9, 10, 11, 12]+"$"  )).indexOf(true); // === 2

在Array中查找子数组

3 个答案: