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houseRobber.cpp
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99 lines (88 loc) · 2.58 KB
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// Source : https://leetcode.com/problems/house-robber/
// Author : Hao Chen
// Date : 2015-04-07
/**********************************************************************************
*
* You are a professional robber planning to rob houses along a street. Each house has
* a certain amount of money stashed, the only constraint stopping you from robbing
* each of them is that adjacent houses have security system connected and it will
* automatically contact the police if two adjacent houses were broken into on the same night.
*
* Given a list of non-negative integers representing the amount of money of each house,
* determine the maximum amount of money you can rob tonight without alerting the police.
*
*
**********************************************************************************/
#include <time.h>
#include <stdlib.h>
#include <iostream>
#include <vector>
using namespace std;
/*
* Dynamic Programming
*
* We can easy find the recurive fomular:
*
* dp[n] = max(
* dp[n-1], // the previous house has been robbed.
* dp[n-2] + money[n] // the previous house has NOT been robbed.
* )
*
* The initalization is obvious:
* dp[1] = money[1]
* dp[2] = max(money[1], money[2])
*
*/
int rob1(vector<int> &money) {
int n = money.size();
if (n==0) return 0;
vector<int> dp(n, 0);
if (n>=1) dp[0] = money[0];
if (n>=2) dp[1] = max(money[0], money[1]);
for (int i=2; i<n; i++){
dp[i] = max(dp[i-1], dp[i-2] + money[i]);
}
return dp[n-1];
}
/*
* Acutally, we no need to allocate an additional array for DP.
* we can only use several variables to record previous steps
*/
int rob2(vector<int> &money) {
int n2=0; // dp[i-2];
int n1=0; // dp[i-1];
for (int i=0; i<money.size(); i++){
int current = max(n1, n2 + money[i]);
n2 = n1;
n1 = current;
}
return n1;
}
int rob(vector<int> &num) {
if (rand()%2)
return rob1(num);
return rob2(num);
}
void printVector( vector<int> &v ){
cout << '[' ;
for(int i=0; i<v.size(); i++){
cout << v[i] << (i==v.size()-1 ? " " :", ");
}
cout << ']' << endl;
}
int main(int argc, char** argv) {
srand(time(0));
vector<int> money;
if (argc>1){
for (int i=1; i<argc; i++) {
money.push_back(atoi(argv[i]));
}
}else{
money.push_back(2);
money.push_back(1);
money.push_back(3);
money.push_back(4);
}
printVector(money);
cout << rob(money) << endl;
}