Missing Number — Leetcode
1 min readJul 29, 2022
Problem statement — 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.
1.XOR
public int missingNumber(int[] nums) { //xor
int res = nums.length;
for(int i=0; i<nums.length; i++){
res ^= i;
res ^= nums[i];
}
return res;
}
2.SUM
public int missingNumber(int[] nums) { //sum
int len = nums.length;
int sum = (0+len)*(len+1)/2;
for(int i=0; i<len; i++)
sum-=nums[i];
return sum;
}
3.Binary Search
public int missingNumber(int[] nums) { //binary search
Arrays.sort(nums);
int left = 0, right = nums.length, mid= (left + right)/2;
while(left<right){
mid = (left + right)/2;
if(nums[mid]>mid) right = mid;
else left = mid+1;
}
return left;
}
Summary:
If the array is in order, I prefer Binary Search
method. Otherwise, the XOR
method is better