这道题目可以使用一般的DP来做，但是博主想到的方法更加直接：对于判断是否存在解，直接把数组累加，如果和非负，那么必然存在一个位置，使得从这个位置开始进行累加的所有结构都非负。在进行累加的过程中，找到累加和最小的位置，从这个位置的下一个位置开始，就可以保证在累加过程中，累加曲线的最小值最大（想象积分曲线的过程，积分的终点sum相同，一个先下降再上升，一个先上升再下降）。有点绕，多想想 ：)
题目
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- Difficulty: Medium
- Contributor: LeetCode
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.
代码
分析就在摘要里面了，直接上代码。
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
const int n = gas.size();
int cum = 0, min = INT_MAX, minid = -1;
for (int i = 0; i < n; ++i){
int tmp = gas[i] - cost[i];
cum += tmp;
if (cum < min){
min = cum;
minid = i + 1;
}
}
if (minid == n){
minid = 0;
}
return (cum < 0) ? -1 : minid;
}
};