标签：code hat array tin while binary mis number linear
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?
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;
}
标签：code hat array tin while binary mis number linear
原文地址：http://www.cnblogs.com/apanda009/p/7643717.html
