2015-10-18 10:42:52 +08:00

178 lines
4.0 KiB
C++

// Source : https://oj.leetcode.com/problems/first-missing-positive/
// Author : Hao Chen
// Date : 2014-07-18
/**********************************************************************************
*
* 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.
*
*
**********************************************************************************/
#include <stdlib.h>
#include <time.h>
#include <iostream>
#include <map>
using namespace std;
#define INT_MAX 2147483647
/*
* Idea:
*
* We can move the num to the place whcih the index is the num.
*
* for example, (considering the array is zero-based.
* 1 => A[0], 2 => A[1], 3=>A[2]
*
* Then, we can go through the array check the i+1 == A[i], if not ,just return i+1;
*
* This solution comes from StackOverflow.com
* http://stackoverflow.com/questions/1586858/find-the-smallest-integer-not-in-a-list
*/
int firstMissingPositive_move(int A[], int n) {
if (n<=0) return 1;
int num;
for(int i=0; i<n; i++) {
num = A[i];
while (num>0 && num<n && A[num-1]!=num) {
swap(A[i], A[num-1]);
num = A[i];
}
}
for (int i=0; i<n; i++){
if (i+1 != A[i]){
return i+1;
}
}
return n+1;
}
/*
* The idea is simple:
*
* 1) put all of number into a map.
* 2) for each number a[i] in array, remove its continous number in the map
* 2.1) remove ... a[i]-3, a[i]-2, a[i]-1, a[i]
* 2.2) remove a[i]+1, a[i]+2, a[i]+3,...
* 3) during the removeing process, if some number cannot be found, which means it's missed.
*
* considering a case [-2, -1, 4,5,6],
* [-2, -1] => missed 0
* [4,5,6] => missed 3
*
* However, we missed 1, so, we have to add dummy number 0 whatever.
*
* NOTE: this solution is not constant space slution!!!!
*
*/
int firstMissingPositive_map(int A[], int n) {
map<int, int> cache;
for(int i=0; i<n; i++){
cache[A[i]] = i;
}
//add dummy
if (cache.find(0)==cache.end() ) {
cache[0] = -1;
}
int miss = INT_MAX;
int x;
for (int i=-1; i<n && cache.size()>0; i++){
if (i == -1){
x = 0; //checking dummy
}else{
x = A[i];
}
if ( cache.find(x)==cache.end() ){
continue;
}
int num ;
// remove the ... x-3, x-2, x-1, x
for( num=x; cache.find(num)!=cache.end(); num--){
cache.erase(cache.find(num));
}
if ( num>0 && num < miss ){
miss = num;
}
// remove the x+1, x+2, x+3 ...
for ( num=x+1; cache.find(num)!=cache.end(); num++){
cache.erase(cache.find(num));
}
if ( num>0 && num < miss) {
miss = num;
}
}
return miss;
}
int firstMissingPositive(int A[], int n) {
srand(time(0));
if (rand()%2){
return firstMissingPositive_move(A, n);
}
return firstMissingPositive_map(A, n);
}
void printArray(int a[], int n){
cout << "[ ";
for(int i=0; i<n-1; i++) {
cout << a[i] << ", ";
}
cout << a[n-1] << " ]";
}
void Test(int a[], int n, int expected) {
printArray(a, n);
int ret = firstMissingPositive(a, n);
cout << "\t missed = " << ret << " " << (ret==expected?"passed!":"failed!") << endl;
//printArray(a, n);
//cout <<endl;
}
int main()
{
#define TEST(a, e) Test(a, sizeof(a)/sizeof(int), e)
int a0[]={1};
TEST(a0, 2);
int a1[]={1,2,0};
TEST(a1, 3);
int a2[]={3,4,-1,1};
TEST(a2, 2);
int a3[]={1000,-1};
TEST(a3, 1);
int a4[]={1000, 200};
TEST(a4, 1);
int a5[]={2,5,3,-1};
TEST(a5, 1);
int a6[]={1, 100};
TEST(a6, 2);
int a7[]={7,8,9,11};
TEST(a7, 1);
int a8[]={4,3,2,1};
TEST(a8, 5);
return 0;
}