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.

Example 1

Input: [3,0,1]
Output: 2

Example 2

Input: [9,6,4,2,3,5,7,0,1]
Output: 8

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums.sort()
        n = len(nums)
        for i in range(0, n):
            if nums[i] != i:
                return i
        return n

自己写排序的话：

class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # [3,0,1]
        # 3 != nums[3-1], swap, [1, 0, 3]
        # 1 == nums[1-1]
        # 0 pass
        # 3 == nums[3-1] 
        # [9,6,4,2,3,5,7,0,1]
        for i in xrange(0, len(nums)):            
            while nums[i] != 0 and nums[i] != nums[nums[i]-1]:
                nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1]               
        for i in xrange(0, len(nums)):
            if nums[i] != i+1:
                return i+1
        return 0        

对于1～n的数字排序，直接O（n）可以搞定哇！

利用数学知识搞定：

class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = len(nums)
        return n*(n+1)/2-sum(nums)        

class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        xor = len(nums)
        for i,n in enumerate(nums):
            xor = xor^n^i
        return xor

还有使用二分搞定：

class Solution {
    // 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;
    }

}