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？

Li Shi's Analysis:

The idea is to use the array itself as a table. We try put a value i on the place i-1. If the value i is out of range, i.e., <=0 || >n, then ignore it. At each position i, we check whether the A[i] is a value needed to be placed on its own position, if yes, we swap it; then we check the new A[i] again. We repeat this procedure until A[i] is a value that is out of range or the right place of value A[i] is already placed a value of A[i] (Duplicate situation), we then move the i+1 position.

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

public class Solution {
    public int firstMissingPositive(int[] A) {
        if (A==null && A.length==0) {
            return 1;
        }
        for (int i=0; i<A.length; i++) {
            if (A[i]>0 && A[i]<=A.length && A[i]!=A[A[i]-1]) {
                int temp = A[A[i]-1];
                A[A[i]-1] = A[i];
                A[i] = temp;
                i--;
            }
        }
        
        for (int i=0; i<A.length; i++) {
            if (A[i] != i+1) return i+1;
        }
        return A.length+1;
    }
}

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

public class Solution {
    public int firstMissingPositive(int[] A) {
        int shouldbe = 1;
        int store = 0;
        java.util.Arrays.sort(A);
        for (int elem : A) {
            if (elem > 0) {
                if (elem == store) continue;
                else if (elem == shouldbe) {
                    shouldbe++;
                    store = elem;
                }
                else break;
            }
        }
        return shouldbe;
    }
}