0268. Missing Number (E)

Missing Number (E)

题目

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?

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题意

给定一个长度为n的数组，从 0-n 这 n+1 个数中选取 n 个数填充入数组中，找到缺失的那一个数。

思路

交换法：第一次遍历数组，将nums[i]与下标为nums[i]的数进行交换，第二次遍历数组，如果遇到 i != nums[i]，说明i就是缺失的数。

求和法：先求 0-n 的和，再求数组中数的和，相减就是缺失的数。

异或法：根据相同的数异或为0的性质， 将数组中所有数异或一遍，再将结果与 0-n 异或一遍，相同的数会互相抵消，得到的结果就是缺失的数。

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代码实现

Java

交换

class Solution {
    public int missingNumber(int[] nums) {
        int i = 0;
        while (i < nums.length) {
            if (nums[i] >= nums.length || nums[i] == i) {
                i++;
            } else {
                swap(nums, i, nums[i]);
            }
        }
        for (int j = 0; j < nums.length; j++) {
            if (nums[j] != j) {
                return j;
            }
        }
        return nums.length;
    }

    private void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
}

求和

class Solution {
    public int missingNumber(int[] nums) {
        int len = nums.length;
        int sum1 = (len + 1) * len / 2;
        int sum2 = 0;
        for (int num : nums) {
            sum2 += num;
        }
        return sum1 - sum2;
    }
}

异或

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