"Leetcode" First Missing Positive (2 solutions)

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.

Solution One: O (nlogn) Time and O (1) space

No brain solution--First sort O (Nlogn)

Thoughts are as follows:

1, omit the number of non-positive prefixes.

2. Write down a positive integer to be matched as tag.

Consider repetition, if a[i] equals tag, then tag++

If A[i] is greater than tag, returns the tag

A[i] cannot be less than tag, which can be guaranteed by sequencing.

classSolution { Public:    intFirstmissingpositive (intA[],intN) {if(n = =0)            return 1; Sort (A,a+N); inti =0;  while(I < n && A[i] <=0) I++; if(i = =N)return 1; intTag =1;  for(; i < n; i + +))        {            if(A[i] >tag)//miss the Tag                returntag; Else if(A[i] = =tag)//Next PositiveTag + +; Else                ; }        //I==n, miss the tag        returntag; }};

Solution two: O (n) time and O (n) space

A little thought would tell that an array of n capacity, with a maximum of the positive integers covered, is a continuous 1~n

In other words, a missing positive integer is either present in 1~n or n+1

You can therefore construct an array of size n Tag,tag[i] to record whether i+1 this number appears in a.

Returns N+1 if tag is true, otherwise returns the first tag with a negative label of +1

classSolution { Public:    intFirstmissingpositive (intA[],intN) {if(n = =0)            return 1; //Tag[i] means whether i+1 exists in A//at the most 1~n and then return n+1vector<BOOL> tag (n,false);  for(inti =0; I < n; i + +)        {            if(A[i] >0&& A[i] <=N) tag[a[i]-1] =true; }         for(inti =0; I < n; i + +)        {            if(Tag[i] = =false)                returni+1; }        returnn+1; }};

"Leetcode" First Missing Positive (2 solutions)