Leetcode: First Missing Positive

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



For example,

Given [1,2,0] return 3,

and [3,4,-1,1] return 2.



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

Naive的方法肯定是sort这个数组，但那样要用O(NlongN)的时间；第二种思路是用哈希表映射，Mapping all positive integers to a hash table and iterate from 1 to the length of the array to find out the first missing one，但哈希表是O(N)的space. 那怎么做呢？

不能开辟非常数的额外空间，就需要在原数组上操作，思路是交换数组元素，让数组中index为i的位置存放数值(i+1)。最后如果哪个数组元素违反了A[i]=i+1即说明i+1就是我们要求的第一个缺失的正数。具体操作如下：

• if A[i] is positive, say we have A[i] = x, we know it should be in slot A[x-1]! That is to say, we can swap A[x-1] with A[i] so as to place x into the right place.
• if A[i] is non-positive, 0, or A[i]>A.length, ignore it;
• if A[i] == A[A[i]-1], we do not swap, two cases, 1. Maybe A[i] = i+1, A[i]is correct; 2. though A[i] is not correct like [1, 1] index 1, but A[0] == A[1], swap A[1] with A[0] will only give rise to infinite loop

 1 public class Solution {

 2     public int firstMissingPositive(int[] A) {

 3         if (A==null && A.length==0) {

 4             return 1;

 5         }

 6         for (int i=0; i<A.length; i++) {

 7             if (A[i]>0 && A[i]<=A.length && A[i]!=A[A[i]-1]) {

 8                 int temp = A[A[i]-1];

 9                 A[A[i]-1] = A[i];

10                 A[i] = temp;

11                 i--;

12             }

13         }

14         

15         for (int i=0; i<A.length; i++) {

16             if (A[i] != i+1) return i+1;

17         }

18         return A.length+1;

19     }

20 }

有个细节是8-10行交换两个元素，不能先改动A[i]，这样A[A[i]-1]会变。结果就是无限循环。一定要先改A[A[i]-1]。

实现中还需要注意一个细节，第7行，一定是A[i]跟A[A[i]-1]比，而不能是跟i+1比（会出现无限循环，比如输入[1, 1]）。就是如果当前的数字所对应的下标已经是对应数字了，那么我们也需要跳过，因为那个位置的数字已经满足要求了，否则会出现一直来回交换的死循环。这样一来我们只需要扫描数组两遍，时间复杂度是O(2*n)=O(n)，而且利用数组本身空间，只需要一个额外变量，所以空间复杂度是O(1)。

 

Naive方法，时间复杂度O(NlogN), Analysis: 第一次写的时候没有考虑到数组元素有可能重合的问题

 1 public class Solution {

 2     public int firstMissingPositive(int[] A) {

 3         int shouldbe = 1;

 4         int store = 0;

 5         java.util.Arrays.sort(A);

 6         for (int elem : A) {

 7             if (elem > 0) {

 8                 if (elem == store) continue;

 9                 else if (elem == shouldbe) {

10                     shouldbe++;

11                     store = elem;

12                 }

13                 else break;

14             }

15         }

16         return shouldbe;

17     }

18 }