41. First Missing Positive


Given an unsorted integer array nums, return the smallest missing positive integer.

You must implement an algorithm that runs in O(n) time and uses O(1) auxiliary space.

Example 1:

Input: nums = [1,2,0]
Output: 3
Explanation: The numbers in the range [1,2] are all in the array.

Example 2:

Input: nums = [3,4,-1,1]
Output: 2
Explanation: 1 is in the array but 2 is missing.

Example 3:

Input: nums = [7,8,9,11,12]
Output: 1
Explanation: The smallest positive integer 1 is missing.

Constraints:

  • 1 <= nums.length <= 10^5
  • -2^31 <= nums[i] <= 2^31 - 1

这道题让我们找缺失的首个正数,由于限定了 O(n) 的时间,所以一般的排序方法都不能用,最开始博主没有看到还限制了空间复杂度,所以想到了用 HashSet 来解,这个思路很简单,把所有的数都存入 HashSet 中,然后循环从1开始递增找数字,哪个数字找不到就返回哪个数字,如果一直找到了最大的数字(这里是 nums 数组的长度),则加1后返回结果 res,参见代码如下:

解法一:

// NOT constant space
class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        unordered_set<int> st(nums.begin(), nums.end());
        int res = 1, n = nums.size();
        while (res <= n) {
            if (!st.count(res)) return res;
            ++res;
        }
        return res;
    }
};

但是上面的解法不是 O(1) 的空间复杂度,所以需要另想一种解法,既然不能建立新的数组,那么只能覆盖原有数组,思路是把1放在数组第一个位置 nums[0],2放在第二个位置 nums[1],即需要把 nums[i] 放在 nums[nums[i] - 1]上,遍历整个数组,如果 nums[i] != i + 1, 而 nums[i] 为整数且不大于n,另外 nums[i] 不等于 nums[nums[i] - 1] 的话,将两者位置调换,如果不满足上述条件直接跳过,最后再遍历一遍数组,如果对应位置上的数不正确则返回正确的数,参见代码如下:

解法二:

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

Github 同步地址:

https://github.com/grandyang/leetcode/issues/41

类似题目:

Missing Number

Find the Duplicate Number

Find All Numbers Disappeared in an Array

Couples Holding Hands

Smallest Number in Infinite Set

Maximum Number of Integers to Choose From a Range I

Smallest Missing Non-negative Integer After Operations

Maximum Number of Integers to Choose From a Range II

参考资料:

https://leetcode.com/problems/first-missing-positive/

https://leetcode.com/problems/first-missing-positive/discuss/17071/My-short-c++-solution-O(1)-space-and-O(n)-time

LeetCode All in One 题目讲解汇总(持续更新中…)

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