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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]
题目链接：https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/
题目分析：有点像First Missing Positive，思想就是将数字i放到第i-1位上
6ms，时间击败92.45%，空间击败94.76%
class Solution {public void swap(int[] nums, int i, int j) {int tmp = nums[i];nums[i] = nums[j];nums[j] = tmp;}public List<Integer> findDisappearedNumbers(int[] nums) {List<Integer> ans = new ArrayList<>();int pos = 0;while (pos < nums.length) {while (nums[pos] != pos + 1 && nums[pos] != nums[nums[pos] - 1]) {swap(nums, pos, nums[pos] - 1);}pos++;}for (int i = 0; i < nums.length; i++) {if (nums[i] != i + 1) {ans.add(i + 1);}}return ans;}
}// 4 3 2 7 8 2 3 1
// 7 3 2 4 8 2 3 1
// 3 3 2 4 8 2 7 1
// 2 3 3 4 8 2 7 1
// 3 2 3 4 8 2 7 1
// 3 2 3 4 1 2 7 8
// 1 2 3 4 3 2 7 8
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