Hard
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 constant extra space.
Example 1:
Input: nums = [1,2,0]
Output: 3
Example 2:
Input: nums = [3,4,-1,1]
Output: 2
Example 3:
Input: nums = [7,8,9,11,12]
Output: 1
Constraints:
1 <= nums.length <= 5 * 105
-231 <= nums[i] <= 231 - 1
class Solution {
fun firstMissingPositive(nums: IntArray): Int {
var noOne = true
for (i in 0 until nums.size) {
if (noOne && nums[i] == 1) {
noOne = false
} else if (nums[i] <= 0) {
nums[i] = 1
}
}
if (noOne) {
return 1
}
var high = 0
var k: Int
for (x in nums) {
k = kotlin.math.abs(x)
high = kotlin.math.max(high, k)
if (k - 1 < nums.size) {
nums[k - 1] = -1 * kotlin.math.abs(nums[k - 1])
}
}
for (i in nums.indices) {
if (nums[i] > 0) {
return i + 1
}
}
return high + 1
}
}