problem：
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 ⑴.
Note:
The solution is guaranteed to be unique.
Hide Tags
Greedy
题意：汽车从某1个加油站动身，加油、消耗，找出唯1那个可以回到原点的加油站
thinking：
（1）算法比较简单，汽油的余量为负数时行不通，贪心策略
（2）我刚开始采取的DFS+剪分支的方式，提交超时，但这个思路适用面更广
（3）采取数组处理的方式可以免函数递归调用的开消
code：
DFS+剪分支：超时
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int dep=0;
int maxDep=gas.size()⑴;
if(maxDep<0)
return 0;
int fuel=0;
bool flag=true;
for(int index=0;index<gas.size();index++)
{
dfs(dep,maxDep,index,fuel,gas,cost,flag);
if(flag)
return index;
flag=true;
}
return ⑴;
}
protected:
void dfs(int dep,int maxDep,int index,int fuel, vector<int> &gas, vector<int> &cost, bool &flag)
{
if(!flag)
return;
fuel+=gas[index];
fuel-=cost[index];
if(fuel<0)
{
flag=false;
return;
}
if(dep==maxDep)
{
if(fuel<0)
flag=false;
else
flag=true;
return;
}
dfs(dep+1,maxDep,(index+1)%gas.size(),fuel,gas,cost,flag);
}
};
贪心策略：AC
时间复杂度O（n）
class Solution {
public:
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
int total = 0;
int j = ⑴;
for (int i = 0, sum = 0; i < gas.size(); ++i) {
sum += gas[i] - cost[i];
total += gas[i] - cost[i];
if (sum < 0) {
j = i;
sum = 0;
}
}
return total >= 0 ? j + 1 : ⑴;
}
};