268. Missing Number

最新推荐文章于 2023-10-11 12:12:13 发布

原创最新推荐文章于 2023-10-11 12:12:13 发布 · 106 阅读

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本文介绍了一种在给定数组中找到缺失数字的有效算法。该算法通过原地交换元素达到线性时间复杂度O(n)且空间复杂度O(1)。此外还探讨了排序、求和及位运算等其他解决方案。

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?

今天再认真看这道题才发现之前的解法是错误的

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

排序的时间复杂度是O(nlgn)，而且Arrays.sort的空间复杂度也不是O(1)

修改为如下代码，时间复杂度O(n) (每个数字都会被最多操作一次，即放置到正确的位置上)

    public int missingNumber(int[] nums) {
        for (int i=0; i<nums.length; i++) {
            while (nums[i] != i && nums[i] != nums.length) {
                swap(nums, i, nums[i]);
            }
        }
        for (int i=0; i<nums.length; i++) {
            if (nums[i] != i) {
                return i;
            }
        }
        return nums.length;
    }
    
    private void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }

看了discuss，还有几种解法

一是二分查找，但是这需要保证输入是有序的

二是先算sum=n(n+1)/2，然后逐个做差，剩下的就是missing num，但是如果数组的数字比较大，求和会overflow

最后一种是位运算，比较好，如下

The basic idea is to use XOR operation. We all know that a^b^b =a, which means two xor operations with the same number will eliminate the number and reveal the original number.

In this solution, I apply XOR operation to both the index and value of the array. In a complete array with no missing numbers, the index and value should be perfectly corresponding( nums[index] = index), so in a missing array, what left finally is the missing number.

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;
}