LeetCode Missing Number

原题链接在这里：https://leetcode.com/problems/missing-number/

题目：

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?

再刷时，感觉好极了！

与First Missing Positive相似。第一遍把nums[i]放到对应位置上，第二遍看谁不在对应位置上。

Time Complexity: O(n). Space: O(1).

AC Java:

public class Solution {
public int missingNumber(int[] nums) {
if(nums == null || nums.length == 0){
return 0;
}
for(int i = 0; i<nums.length; i++){
if(nums[i] >= 0 && nums[i] < nums.length && nums[i] != nums[nums[i]]){
swap(nums, i, nums[i]);
i--;
}
}
for(int i = 0; i<nums.length; i++){
if(nums[i] != i){
return i;
}
}
return nums.length;
}
private void swap(int [] nums, int i, int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}

第一种方法是 求和，然后挨个减掉，剩余的值就是结果，因为如果是全的，那么sum = n*(n+1)/2.

第二种方法是从前往后做bit Manipulation, res初始为0，每次保留结果 和 (i+1)^nums[i] 取结果，方法非常巧妙。

AC Java:

public class Solution {
public int missingNumber(int[] nums) {
/*
//Method 1
if(nums == null || nums.length == 0){
return 0;
}
int n = nums.length;
int sum = 0;
for(int i = 0; i<nums.length; i++){
sum+=nums[i];
}
return n*(n+1)/2 - sum;
*/
//Method 2
if(nums == null || nums.length == 0){
return 0;
}
int res = 0;
for(int i = 0; i<nums.length; i++){
res ^= (i+1)^nums[i];
}
return res;
}
}