[LeetCode] 268. Missing Number 缺失的数字

Posted 2020-10-28 轻风舞动

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

给n个数，0到n之间有一个数字缺失了，让找到这个缺失数字。要求线性时间复杂度和常数级的额外空间复杂度。

解法1：排序，然后二分查找。此题数组不是排序的，所以适合。但面试时，如果问道要知道。

解法2：求差值，用0~n个数的和减去给定数组数字的和，差值就是要找的数字。

解法3：位操作Bit Manipulation，将这个数组与0~n之间完整的数异或一下累加到res，相同的数字异或后都变为了0，最后剩下的结果就是缺失的数字。

Java:

class Solution {
    public int missingNumber(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int result = 0;

        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            if (result != nums[i]) {
                return result;
            }
            result++;
        }
        return nums.length;
    }
}　　

Java:

class Solution {
    public int missingNumber(int[] nums) {
        int xor = 0, i = 0;

        for (i = 0; i < nums.length; i++) {
            xor = xor ^ i ^ nums[i];
        }

        return xor ^ i;
    }
}　　

Python:

class Solution(object):
    def missingNumber(self, nums):
        return sum(xrange(len(nums)+1)) - sum(nums)

Python:

class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return reduce(operator.xor, nums,                       reduce(operator.xor, xrange(len(nums) + 1)))　　

C++: 用等差数列公式

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int sum = 0, n = nums.size();
        for (auto &a : nums) {
            sum += a;
        }
        return 0.5 * n * (n + 1) - sum;
    }
};

C++:

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int res = 0;
        for (int i = 0; i < nums.size(); ++i) {
            res ^= (i + 1) ^ nums[i];
        }
        return res;
    }
};

 

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