Gas Station
来源:互联网 发布:centos安装新内核 编辑:程序博客网 时间:2024/06/10 17:46
思路:
模拟。如果途经某一点,使得sum变负,说明刚经过的那点cost太大,应该考虑从下一点重新开始。
时间复杂度O(N), 空间复杂度O(1)。
class Solution {public: int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { int sum = 0, total = 0, j = -1; for(int i = 0; i < gas.size(); ++i) { sum += gas[i] - cost[i]; total += gas[i] - cost[i]; if(sum < 0) { sum = 0; j = i; } } return total >= 0 ? j+1 : -1; }};
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