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.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
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.
Solution
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
int firstMissingPositive(vector<int>& nums) {
int len = nums.size();
for (int i = 0 ; i < len; i++) {
if (nums[i] <= 0) {
nums[i] = len + 1;
}
}
for (int i = 0; i < len; i++) {
int num = abs(nums[i]);
if (num <= len) {
nums[num - 1] = -abs(nums[num - 1]);
}
}
for (int i = 0; i < len; i++) {
if (nums[i] > 0) {
return i + 1;
}
}
return len + 1;
}
