Question

如果我有一个列表l1=["A","B","C"]，并且我有另一个列表l2=["X","Y","Z","A","B","C","1","2","3"]，我该如何有效地检查l1中是否还出现了l2，还有订单。< / p>

在这种情况下，它会显示在列表中，但如果l1为["A","C","B"]，则不会。或者，如果它是l2=["X","Y","Z","A","D","B", "C","1","2","3"]，也应该找不到它，因为它也必须是顺序的。

由于

Answer 1

＆＃13;

function check(l1, l2) {
	return !!~l2.join('').indexOf(l1.join(''))
}
console.log(check(["A","B","C"], ["X","Y","Z","A","B","C","1","2","3"]));
console.log(check(["A","B","C"], ["X","Y","Z","A","D","B", "C","1","2","3"]));

＆＃13;

<强>更新

这是如何运作的：

l1.join('')和l2.join('')将数组转换为字符串 join on MDN

.indexOf告诉我们l2 indexOf on MDN

中l1字符串的毒药在哪里？

~技巧是一个Bitwise NOT运算符而没有进入细节，它在Java脚本中将-1转换为0并将0转换为-1除true

外，每个号码均为0

!!（双重否定）将Number类型转换为Boolean类型What is the !! (not not) operator in JavaScript?

Answer 2

如果在第二个数组中找到，可以使用every()方法并存储第一个元素的索引，然后只增加该索引并检查它是否存在于第二个数组中。

＆＃13;

var l1 = ["A", "B", "C"]
var l2 = ["X", "Y", "Z", "A", "B", "C", "1", "2", "3"]

var check = l1.every(function(e, i) {
  if (i == 0) {
    var start = l2.indexOf(e);
    if (start == -1) return false;
    else {
      this.start = start;
      return true;
    }
  } else {
    return e == l2[this.start += 1]
  }
}, {});

console.log(check)

＆＃13;

更新：您还可以使用every()和some()。

＆＃13;

var l1 = ["A", "B", "C"]
var l2 = ["X", "A", "Z", "A", "B", "C", "1", "2", "3"]

function check(arr1, arr2) {
  return arr2.some(function(e, i) {
    if (e == arr1[0]) {
      return arr1.every((a, j) => a == arr2[i + j])
    }
  })
}


console.log(check(l1, l2))

＆＃13;

Answer 3

我用过它，似乎工作正常。有没有人看到任何问题？

        var l1 = ["A", "B", "C"]
        var l2 = ["X", "Y", "Z", "a", "B", "C", "1", "2", "3"]


        console.log(search(l1,l2,0));


        function CIindexOf(lst, val, offset) {
            var lVal = val.toLowerCase();
            for(var i=offset; i<lst.length; i++) {
                if (lst[i].toLowerCase() == lVal) return i;
            }
            return -1;
        }
        function search(l1, l2, offset) {
            if (l1.length == 0) return 0;
            var index = CIindexOf(l2, l1[0], offset);
            if (index == -1) return -1;
            for (var i=1; i<l1.length; i++) {
                if (l1[i].toLowerCase() != l2[index+i].toLowerCase()) return search(l1, l2, index+1);
            }
            return index;
        }

Answer 4

此版本具有良好的性能（方法中没有回调），可以处理任何类型的内容。

function isContain(arr1, arr2) {
	var arr2Length = arr2.length;
	if (arr2Length > arr1.length) return false; // quick check if arr2 is bigger than arr1 is always false
	var lastPos = -1;
	var i;
	while (~(lastPos = arr1.indexOf(arr2[0], lastPos + 1))) {
		for (i = 0; i < arr2Length; i++) {
			if (arr1[lastPos + i] !== arr2[i]) break;
		}
		if (i === arr2Length ) return true;
	}
	return false;
}

console.log(isContain(["X","Y","Z","A","B","C","1","2","3"], ["A","B","C"]));
console.log(isContain(["X","Y","Z","A","D","B", "C","1","2","3"], ["A","B","C"]));
console.log(isContain(["X","Y","Z","A","B","C","1","2","3"], ["AB","C"]));
console.log(isContain(["X","Y","Z","AB","C","1","2","3"], ["AB","C"]));
console.log(isContain([false, false, true, false, false, true], [true, true]));
console.log(isContain([false, false, true, false, false, true], [true, false]));
console.log(isContain([false, false, true, false, false, true], [1, 0]));
console.log(isContain([1, 1], [1, 1, 1]));

如何检查列表是否是另一个包含订单的列表的子列表？

4 个答案: