Missing Positive (MAP)

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.

Idea: N number, the possible maximum positive integer is n, so you can use this n number as the hash value. But there is an extra space for O (n).

The solution is to reuse a[], first a[i] negative numbers are removed (marked as n+1), and secondly, when traversing to an integer cur, the a[cur] becomes the opposite number, so that the a[i]<0 I position I appeared, but also to ensure that the original value is not destroyed (can be obtained by ABS (A[i]) obtained.

classSolution { Public:    intFirstmissingpositive (intA[],intN) {inti;  for(i =0; I < n; i++)            if(a[i]<=0) A[i] = n+1;  for(i =0; I < n; i++) {            if(ABS (A[i]) <= N) {//ABS is required because in the FOR loop we change the value of [1,n] to a negative number                intCur = ABS (A[i])-1; A[cur]= -ABS (A[cur]); }        }         for(i =0; I < n; i++) {            if(A[i] >0)returni+1; }        returnn+1; }};

Missing Positive (MAP)