268. Missing Number

最新推荐文章于 2023-10-11 12:12:13 发布

zshouyi

最新推荐文章于 2023-10-11 12:12:13 发布

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本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/54772274

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本文介绍了一种在给定数组中寻找缺失数字的算法。数组包含从0到n的n个不同数字，需要找到缺失的那个数字。文章提供了三种解决方法：XOR、SUM和BinarySearch，并附带了相应的Java代码实现。

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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?

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

有三种方法可以解题，XOR, SUM, Binary Search。代码如下：

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;
}