leetcode 41|268. First Missing Positive 268. Missing Number

最新推荐文章于 2023-06-08 18:03:11 发布

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最新推荐文章于 2023-06-08 18:03:11 发布

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本文介绍了一种在未排序整数数组中查找首个缺失正整数的算法，通过原地哈希的方式实现了O(n)的时间复杂度和常数级的空间复杂度。同时探讨了寻找从0到n中缺失数字的问题。

41. First Missing Positive

Given an unsorted integer array, find the first missing positive integer.

For example,
Given [1,2,0] return 3,
and [3,4,-1,1] return 2.

Your algorithm should run in O(n) time and uses constant space.

按照nums[i]的值将它移动到该去的位置那就是nums[nums[i]-1],比如说nums[i]=5，那么应该在下标为4的位置

class Solution {
public:
    int firstMissingPositive(vector<int>& nums) 
    {
        int i = 0;
        while (i < nums.size())
        {
            if (nums[i] > 0 && nums[i] <= nums.size() && nums[i] != i + 1 && nums[i] != nums[nums[i]-1])
                swap(nums[i], nums[nums[i]-1]);
            else
                i++;
        }
        for (int k = 0; k < nums.size(); k++)
        {
           if (nums[k] != k + 1)
              return k + 1;
        }
        return nums.size() + 1;
    }
};

268. Missing Number

和上面那个题神似！！

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?

class Solution {
public:
    int missingNumber(vector<int>& nums) 
    {
        for (int i = 0; i < nums.size(); i++)
        {
            if (nums[i] >= nums.size() || nums[i] < 0 || nums[i] == nums[nums[i]])
                continue;
            swap(nums[i], nums[nums[i]]);
            i--;
        }
        int k;
        for (k = 0; k < nums.size(); k++)
        {
            if (nums[k] != k)
                break;
        }
        return k;
    }
};