LeetCode: 268. Missing Number

LeetCode: 268. Missing Number

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?

Credits: Special thanks to @jianchao.li.fighter for adding this

problem and creating all test cases.

自己的答案，13ms：

public class Solution {
public int missingNumber(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
Arrays.sort(nums);
int i = 0;
for (i = 0; i < nums.length; i++) {
if (nums[i] != i) {
return i;
}
}
return i;
}
}

最快的答案，1ms：

public class Solution {
public int missingNumber(int[] nums) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
}

return (nums.length * (nums.length + 1)) / 2 - sum;
}
}

3 different ideas: XOR, SUM, Binary Search. Java code

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;
}

求和：

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;
}

二分查找：

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;
}