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.
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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; } };
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;