Leetcode: Missing Number

转载于 2015-12-24 08:50:00 发布 · 55 阅读

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/EdwardLiu/p/5071932.html

文章标签：

#runtime

本文提供了一种寻找数组中缺失数字的算法，该算法仅需线性时间复杂度，并且在常数空间复杂度下运行。通过将数组元素互换到其正确位置，算法能够定位到缺失的数字。

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Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

因为输入数组是0,1,2,...，n。把nums[i]放到i的位置上. nums[i] != i的即为missing number.

注意6-9行不可先令 temp = nums[i]

 1 public class Solution {
 2     public int missingNumber(int[] nums) {
 3         int res = nums.length;
 4         for (int i=0; i<nums.length; i++) {
 5             if (nums[i] < nums.length && nums[nums[i]] != nums[i]) {
 6                 int temp = nums[nums[i]];
 7                 nums[nums[i]] = nums[i];
 8                 nums[i] = temp;
 9                 i--;
10             }
11         }
12         for (int i=0; i<nums.length; i++) {
13             if (nums[i] != i) res = i;
14         }
15         return res;
16     }
17 }