LeetCode 41. First Missing Positive(Hard)

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Problem

41. First Missing Positive(Hard)

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

Example 1:

Input: [1,2,0]
Output: 3

Example 2:

Input: [3,4,-1,1]
Output: 2

Example 3:

Input: [7,8,9,11,12]
Output: 1

Note:

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

Solution

题目主要抓住两点信息，即可得出解法：

• 所求firstMissingPositive一定在闭区间[1,n+1]，有该观察，可以得到O(n) time, O(n) space的解法
• 可以用输入数组nums来保存元素的访问情况，则可以将O(n) space的方法优化到O(1) space

O(n) time, O(n) space

class Solution
{
public:
    int firstMissingPositive(vector<int> &nums)
    {
        if (nums.empty())
            return 1;
        int n = nums.size(), res = 0;
        vector<bool> visited(n + 2, false);
        for (auto num : nums)
        {
            if (num <= n && num > 0)
            {
                visited[num] = true;
            }
        }
        for (int i = 1; i <= n + 1; ++i)
        {
            if (!visited[i])
            {
                res = i;
                break;
            }
        }
        return res;
    }
};

O(n) time, O(1) space

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

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