比较两个数组，并返回一个新数组，其中任何项仅存在于原始数组中

Question

[“闪长岩”、“安山岩”、“草”、“泥土”、“粉红羊毛”、“死灌木”]，

[“闪长岩”、“安山岩”、“草”、“泥土”、“死灌木”]应返回[“粉红色羊毛”]。

因为“第一个数组即ARR1中不存在粉红羊毛，但它返回的是一个空数组，这段代码对仅有数字的数组很好，但当数组只包含字符串或带数字的字符串时，代码就不起作用了。

function diff(arr1, arr2) {

    var newArray = arr2.concat(arr1);  //first joininng both arrays inn one and storing it in newArray 

    var newestArray = [];

    for (var i=0 ; i<newArray.length ; i++) {  //NOW COMPARING EACH ELEMENT OF  newArray  WITH ARR1 AD ARR2 AND PUSHING NOT SAME VALUES TO newestArray
        if (arr1.indexOf(newArray[i]) == -1) {
            newestArray.push(newArray[i]);

            if (arr2.indexOf(newArray[i]) == -1) {
                newestArray.push(newArray[i]);
            }
        }
    }

    return newestArray.filter(Boolean);   //It is returning an empty arrray but it should return "pink wool"
}

diff(["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"], ["diorite", "andesite", "grass", "dirt", "dead shrub"]);

Answer 1

这个解法有一个线性的方法，有一个计数的对象。

var array1 = ["diorite", "andesite", "grass", "dirt", "pink wool", "dead shrub"],
    array2 = ["diorite", "andesite", "grass", "dirt", "dead shrub"];

function symmetricDifference(setA, setB) {
    var o = {}, result = [];
    function count(i, o) {
        return function (a) {
            o[a] = o[a] || { count: 0, value: a };
            o[a].count += i;
        };
    }

    setA.forEach(count(1, o));
    setB.forEach(count(-1, o));
    Object.keys(o).forEach(function (k) {
        if (o[k].count) {
            o[k].count = Math.abs(o[k].count);
            while (o[k].count--) {
                result.push(o[k].value);
            }
        }
    });
    return result;
}

document.write('<pre>' + JSON.stringify(symmetricDifference(array1, array2), 0, 4) + '</pre>');