LeetCode 448. Find All Numbers Disappeared in an Array

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Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

Input:[4,3,2,7,8,2,3,1]Output:[5,6]

 

代碼：

/** * @param {number[]} nums * @return {number[]} */var findDisappearedNumbers = function(nums) {  let number = [];  for(let i = 0; i < nums.length; i++) {      let index = Math.abs(nums[i]) - 1; //取得index，將數組index位置的數字變為負數，那麼之後遍曆數組的時候，如果有正數，那麼那個正數index+1便是我們所需結果      nums[index] = -Math.abs(nums[index]);  }    for(let i = 0; i < nums.length; i++) {      if(nums[i] > 0 ) {          number.push(i + 1);      }  }  return number;};

 

LeetCode 448. Find All Numbers Disappeared in an Array