【LeetCode】448.Find All Numbers Disappeared in an Array(Easy)解题报告

【LeetCode】448.Find All Numbers Disappeared in an Array(Easy)解题报告

tags: Array

题目地址：https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/#/description

题目描述：

  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.

Examples:

Input:

[4,3,2,7,8,2,3,1]

Output:

[5,6]

Solutions:

正解：

//time : O(n)
//space : O(n)
public class Solution {
public List<Integer> findDisappearedNumbers(int[] nums) {
List<Integer> ret = new ArrayList<Integer>();

for(int i = 0; i < nums.length; i++) {
int val = Math.abs(nums[i]) - 1;
if(nums[val] > 0) {
nums[val] = -nums[val];
}
}

for(int i = 0; i < nums.length; i++) {
if(nums[i] > 0) {
ret.add(i+1);
}
}
return ret;
}
}

  思想就是说比如nums数组有8个数，那么数组为nums[0]~nums[7]，里面的数为1~8，数减一之后一定有下标对应，将对应下标的数取相反数，那么遍历之后正数的下标加一则为数组中未出现的数。

Date：2017年6月2日