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]
代码：
/** * @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，将数组index位置的数字变为负数，那么之后遍历数组的时候，如果有正数，那么那个正数index+1便是我们所需结果 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
原文：http://www.cnblogs.com/gogolee/p/6659194.html