# 41.First-Missing-Positive

## 题目地址

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

## 题目描述

``````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.``````

## 代码

### Approach #1

The key here is to use swapping to keep constant space and also make use of the length of the array, which means there can be at most n positive integers. So each time we encounter an valid integer, find its correct position and swap. Otherwise we continue.

``````class Solution {
public int firstMissingPositive(int[] nums) {

int i = 0, n = nums.length;
while (i < n) {
// If the current value is in the range of (0,length) and it's not at its correct position,
// swap it to its correct position.
// Else just continue;
if (nums[i] >= 0 && nums[i] < n && nums[nums[i]] != nums[i]) {
swap(nums, i, nums[i]);
} else {
i++;
}
}
int k = 1;

// Check from k=1 to see whether each index and value can be corresponding.
while (k < n && nums[k] == k) {
k++;
}

// If it breaks because of empty array or reaching the end. K must be the first missing number.
if (n == 0 || k < n) {
return k;
} else {
// If k is hiding at position 0, K+1 is the number.
return nums[0] == k ? k + 1 : k;
}
}

private void swap(int[] nums, int i, int j) {
int tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
}
}``````

### Approach #2 Index as a hash key

Complexity Analysis

• Time complexity : O(N) since all we do here is four walks along the array of length `N`.

• Space complexity : O(1) since this is a constant space solution.

``````class Solution {
public int firstMissingPositive(int[] nums) {
int n = nums.length;
int contains = 0;
for (int i = 0; i < n; i++) {
if (nums[i] == 1) {
contains++;
break;
}
}
if (contains == 0) {
return 1;
}

if (n == 1) {
return 2;
}

for (int i = 0; i < n; i++) {
if ((nums[i] <= 0) || (nums[i] > n)) {
nums[i] = 1;
}
}

for (int i = 0; i < n; i++) {
int a = Math.abs(nums[i]);
if (a == n) {
nums[0] = - Math.abs(nums[0]);
} else {
nums[a] = - Math.abs(nums[a]);
}
}

for (int i = 1; i < n; i++) {
if (nums[i] > 0) {
return i;
}
}

if (nums[0] > 0) {
return n;
}

return n + 1;
}
}``````

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