[LeetCode] 268. Missing Number

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

Example 1:

Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.

Example 2:

Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.

Example 3:

Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.

Constraints:

n == nums.length
1 <= n <= 104
0 <= nums[i] <= n
All the numbers of nums are unique.

Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

丢失的数字。

给定一个包含 [0, n] 中 n 个数的数组 nums ，找出 [0, n] 这个范围内没有出现在数组中的那个数。

这道题我提供三种思路

1. 数学方法

如果数组是完整的，整个数组的和应该是（首项+末项）x 项数 / 2 →设为A

事实情况的加和设为B，A - B即可得到失去的数字，代码如下

时间O(n)

空间O(1)

JavaScript实现

 1 /**
 2  * @param {number[]} nums
 3  * @return {number}
 4  */
 5 var missingNumber = function(nums) {
 6     let res = 0;
 7     let len = nums.length;
 8     let expected = ((1 + len) * len) / 2;
 9     let calculated = 0;
10     for (let i = 0; i < nums.length; i++) {
11         calculated += nums[i];
12     }
13     res = expected - calculated;
14     return res;
15 };

Java实现

 1 class Solution {
 2     public int missingNumber(int[] nums) {
 3         int n = nums.length;
 4         int total = n * (n + 1) / 2;
 5         int sum = 0;
 6         for (int num : nums) {
 7             sum += num;
 8         }
 9         return total - sum;
10     }
11 }

2. 位运算方法

因为数组长度是n，数组里面的数字是0-n范围内的，也许我们可以将数组num[i]和他们的下标做一个关联，由此想到了位运算的做法。例子，如果数组是排序好的（没排序也行，因为位运算遵循交换律，此处为解释方便）

每个数字num[i]和下标i做异或运算XOR，若两者相同，结果为0。根据题意，应该最后就剩下某个数字M= num[i]。M和0再做异或运算，还是M。所以M就是那个失踪的数字。代码如下，

时间O(n)

空间O(1)

JavaScript实现

 1 /**
 2  * @param {number[]} nums
 3  * @return {number}
 4  */
 5 var missingNumber = function(nums) {
 6     let res = nums.length;
 7     for (let i = 0; i < nums.length; i++) {
 8         res ^= i ^ nums[i];
 9     }
10     return res;
11 };

Java实现

1 class Solution {
2     public int missingNumber(int[] nums) {
3         int res = nums.length;
4         for (int i = 0; i < nums.length; i++) {
5             res ^= i ^ nums[i];
6         }
7         return res;
8     }
9 }

3. 桶排序bucket sort，还是将数字放到它该去的位置上，这样使得数组有序。第二遍扫描的时候就知道哪个数字是缺失的了。

时间O(n)

空间O(1)

Java实现

 1 class Solution {
 2     public int missingNumber(int[] nums) {
 3         for (int i = 0; i < nums.length; i++) {
 4             while (nums[i] >= 0 && nums[i] < nums.length && nums[i] != nums[nums[i]]) {
 5                 int temp = nums[nums[i]];
 6                 nums[nums[i]] = nums[i];
 7                 nums[i] = temp;
 8             }
 9         }
10 
11         for (int i = 0; i < nums.length; i++) {
12             if (nums[i] != i) {
13                 return i;
14             }
15         }
16         return nums.length;
17     }
18 }

CNoodle

Let the code speak for you.

[LeetCode] 268. Missing Number

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