leetcode@ [134] Gas station (Dynamic Programming)

https://leetcode.com/problems/gas-station/

题目：

There are N gas stations along a circular route, where the amount of gas at station i is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.

Return the starting gas station's index if you can travel around the circuit once, otherwise return -1.

Note:
The solution is guaranteed to be unique.

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int n = gas.size();
        if(n == 0) return -1;
        if(n == 1) {
            if(gas[0] - cost[0] >= 0)  return 0;
            return -1;
        }
        
        vector<int> dif; dif.clear();
        for(int i=0;i<n;++i) dif.push_back(gas[i] - cost[i]);
        for(int i=0;i<n-1;++i) dif.push_back(gas[i] - cost[i]);
        
        vector<int> sum(2*n-1, 0);
        vector<int> dp(2*n-1, 0);
        sum[0] = dif[0];
        dp[0] = 1;
        
        for(int i=1;i<2*n-1;++i) {
            if(sum[i-1] < 0) {
                sum[i] = dif[i];
                dp[i] = 1;
            }
            else {
                sum[i] = sum[i-1] + dif[i];
                dp[i] = dp[i-1] + 1;
            }
            
            if(dp[i] == n && sum[i] >= 0) return i-n+1;
        }
        
        return -1;
    }
};

View Code

 

这里正好可以引申一个：环形数组的最大子数组和问题。解决这个问题就是把“环形” 变成 “直线型”。比如: 原来的环形数组为 v[0], v[1], ..., v[n-1]. 可以在v[n-1] 后面继续放置 v[0], v[1], ..., v[n-1]. 然后我们对这个新的数组求解：最大子数组和问题。但是这里也有一个问题，就是求解出来的最大子数组可能长度会超过n，所以我们维护两个变量：sum[i] 和 len[i]。sum[i] 记录以a[i] 为结束位置的最大子数组和，而len[i] 记录以a[i] 为结束位置的最大子数组的长度。最终的答案可以找当len[i] <= n 的时候， sum[i] 最大的值。

 

int res = INT_MIN;
vector<int> sum(n, 0);
vector<int> len(n, 0);

sum[0] <- v[0];      len[0] <- 1;

for(i<-1 TO 2*n) {
     if(sum[i-1] < 0) {
        sum[i] <- v[i]
        len[i] <- 1
     }
     else {
        sum[i] <- sum[i-1] + v[i]
        len[i] <- len[i-1] + 1  
     }
     if(len[i] <= n) res <- max{res, sum[i]);
}

return res;