Question

我被要求从1..N数组找到丢失的号码。

例如，对于数组：let numArr = [2,4,6,8,3,5,1,9,10];，缺少的数字为7

let numArr=[2,4,6,8,3,5,1,9,10];
numArr.sort(function(a,b){  //sort numArr
  return a-b;
});

let newNumArr=[];
for(let i=1;i<=10;i++){
  newNumArr.push(i);
}

for(let i=0;i<newNumArr.length;i++){  //compare with new arr
  if(newNumArr[i] !== numArr[i]){
    console.log('The missing num is:'+newNumArr[i]);  //The missing num is:7
    break;
  }
}

Answer 1

您可以使用MAP和FILTER找出单独数组中缺少的数字

const numArr = [2, 4, 6, 8, 3, 5, 1, 9, 10];
const missingNumberArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10].map(number => {
    if (!numArr.includes(number)) {
        return number;
    }
}).filter(y => y !== undefined);

Answer 2

您可以使用连续n个数之和的简单逻辑n*(n+1)/2。从上面减去数组的总和将得到缺失的数字

＆＃13;

let numArr=[2,4,6,8,3,5,1,9,10];
var sum = numArr.reduce((a,c) => a+c, 0);

// As the array contains n-1 numbers, here n will be numArr.length + 1
console.log(((numArr.length + 1) * (numArr.length + 2))/2 - sum);

＆＃13;

Answer 3

使用.find：

会更容易

function findMissing(input) {
  input.sort((a, b) => a - b);
  const first = input[0];
  return input.find((num, i) => first + i !== num) - 1;
}
console.log(findMissing([2, 4, 6, 8, 3, 5, 1, 9, 10]));
console.log(findMissing([3, 4, 5, 6, 8, 9, 2]));

（请注意，这也适用于从1开始的数组中查找缺失值）

从1..N items数组中找到缺少的项目

3 个答案: