这里的目标是计算以下数组中出现的次数：[2,3,7,9,7,3,2]
例如，如果用户输入7，输出应该是[2,4]，因为7出现在这两个索引中。到目前为止，我所拥有的看起来像这样
var arr1 = [2,3,7,9,7,3,2];
printArray(arr1);
var indexOfNum = findValueInArray(arr1, num);
if (indexOfNum === -1) {
console.log('%d was not found in the array of random integers.', num);
}
else {
console.log('%d was found in the array of random integers at index %d',num, indexOfNum);
}
我的结果是：
arr[0] = 2
arr[1] = 3
arr[2] = 7
arr[3] = 9
arr[4] = 7
arr[5] = 3
arr[6] = 2
7 was found in the array of random integers at index 2
我知道我很接近了，但我不确定我到底忽略了什么。谢谢你们！
试试这个，
var arr1 = [2,3,7,9,7,3,2];
function occurance(array,element){
var counts = [];
for (i = 0; i < array.length; i++){
if (array[i] === element) {
counts.push(i);
}
}
return counts;
}
occurance(arr1, 2); //returns [0,6]
occurance(arr1, 7); //returns [2,4]
function findValueInArray(arr, num) {
var r = [];
for(i = 0; i < arr.length; i++) (arr[i] == num) ? r.push(i) : '';
return r;
}
function printArray(num, indexOfNum) {
console.log('%d was found ... index %s', num, indexOfNum.join(', '));
}
var arr = [2,3,7,9,7,3,2];
var num = 7;
var indexOfNum = findValueInArray(arr, num);
printArray(num, indexOfNum);
你可以这样做：
var arr = [2, 3, 7, 9, 7, 3, 2];
function occurrences(x, a) {
var pos = [];
a.forEach(function(val, i) {
if (x === a[i]) {
pos.push(i);
}
});
return pos;
}
console.log(occurrences(7, arr)); // [2, 4]
console.log(occurrences(5, arr)); // []
console.log(occurrences(2, arr)); // [0, 6]
