[力扣题解]134. 加油站

题目：134. 加油站

思路

贪心法；

代码

暴力法

class Solution {
public:int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {int i, rest, index, size;size = gas.size();for(i = 0; i < size; i++){// 从 i 开始// 油量储备rest = gas[i] - cost[i];index = (i + 1) % size;// 有油且未到达终点while(rest > 0 && index != i){rest = rest + gas[index] - cost[index];index = (index + 1) % size;}if(rest >= 0 && index == i){return i;}}return -1;}
};

本题应该设置了时长限制，所以暴力法超时了；

贪心 Method 1

// 代码随想录: 贪心 Method 1
//
class Solution {
public:int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {int curSum = 0;int min = INT_MAX; // 从起点出发，邮箱里的油量最小值int i, rest;for(i = 0; i < gas.size(); i++){cout << "i : " << i;rest = gas[i] - cost[i]; // 这一天剩下的油curSum += rest;          // 现在的油cout << ", curSum : " << curSum;if(curSum < min){min = curSum;}}// 情况1：跑不了一圈if(curSum < 0){return -1;  }// 情况2：if(min >= 0){return 0;}for(i = gas.size()-1; i >= 0; i--){rest = gas[i] - cost[i];min += rest;if(min >= 0){return i;}}return -1;}
};

贪心 Method 2

// 贪心 Method 2
class Solution {
public:int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {int i;int totalsum = 0, cursum = 0, start = 0;for(i = 0; i < gas.size(); i++){cout << ", i : " << i;cursum += gas[i] - cost[i];totalsum += gas[i] - cost[i];cout << ", cursum : " << cursum;// 没油了, 重来if(cursum < 0){start = i+1; // 题目条件有“如果有解存在唯一解”cursum = 0;cout << ", start : " << start;}cout << endl;}// 没戏了if(totalsum < 0){return -1;}return start;}
};