LeetCodeP134 Gas Station
2017-03-13 12:06
127 查看
There are N gas stations along a circular route, where the amount of gas at station i is
You have a car with an unlimited gas tank and it costs
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.
题目大意:gas[i]代表当前汽油站的汽油量,cost[i]代表从第i站到第i+1站需要耗费的汽油。开始你的汽车没有油,你能否选择一个站出发,然后环绕一圈回到原点。
思路:
详细的解答请参考转自此处
按常规思路来说,循环遍历数组,如果i站出发不行,那就换i+1站。如果从i站出发,在到达第k站之前油箱都不空,但到达k站后油箱空了,说明从i站开始无法绕一圈,但是我们不必从i+1重新遍历,因为从i站到k站油箱会空,那么从i到k站之间的任何一个站都到不了k站。所以我们选择从k+1站开始遍历。如下:
使用一个数组diff[]记录当前站汽油量和消耗量的差值:diff[i] = gas[i]-cost[i]。
假设从第一站开始到第p站油箱变空,那么sum1=diff[0]+diff[1]+…+diff[p]<0.
此时,按照我们的思路应该从p+1站开始走,假设从p+1到q站油箱又空了,即sum2=diff[p+1]+diff[p+2]+…+diff[q]<0.
那么我们再从q+1开始,假设从q+1到未开始循环的最后一个站时油箱不空,那么sum3=diff[q+1]+diff[q+2]+diff[size-1] > 0.此时我们看能否开会q站,只需看sum3从diff[0]加到diff[p]的过程中是否一直非负。因此我们只需看sum1+sum2+sum3是否大于等于0.而sum1+sum2+sum3就是整个diff数组的累加和,所以如果diff数组累加和大于等于零,就一定存在一个元素,从这个元素出发,绕数组一圈,能保证累加和一直非负。
代码:
public class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int index = 0;
int total = 0, sum = 0;
if (gas.length == 0 || gas.length != cost.length) {
return -1;
}
for (int i = 0; i < gas.length; i++) {
total += gas[i] - cost[i];
if (sum < 0) {
sum = gas[i] - cost[i];
index = i;
} else {
sum += gas[i] - cost[i];
}
}
return total >= 0 ? index : -1;
}
}
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.
题目大意:gas[i]代表当前汽油站的汽油量,cost[i]代表从第i站到第i+1站需要耗费的汽油。开始你的汽车没有油,你能否选择一个站出发,然后环绕一圈回到原点。
思路:
详细的解答请参考转自此处
按常规思路来说,循环遍历数组,如果i站出发不行,那就换i+1站。如果从i站出发,在到达第k站之前油箱都不空,但到达k站后油箱空了,说明从i站开始无法绕一圈,但是我们不必从i+1重新遍历,因为从i站到k站油箱会空,那么从i到k站之间的任何一个站都到不了k站。所以我们选择从k+1站开始遍历。如下:
使用一个数组diff[]记录当前站汽油量和消耗量的差值:diff[i] = gas[i]-cost[i]。
假设从第一站开始到第p站油箱变空,那么sum1=diff[0]+diff[1]+…+diff[p]<0.
此时,按照我们的思路应该从p+1站开始走,假设从p+1到q站油箱又空了,即sum2=diff[p+1]+diff[p+2]+…+diff[q]<0.
那么我们再从q+1开始,假设从q+1到未开始循环的最后一个站时油箱不空,那么sum3=diff[q+1]+diff[q+2]+diff[size-1] > 0.此时我们看能否开会q站,只需看sum3从diff[0]加到diff[p]的过程中是否一直非负。因此我们只需看sum1+sum2+sum3是否大于等于0.而sum1+sum2+sum3就是整个diff数组的累加和,所以如果diff数组累加和大于等于零,就一定存在一个元素,从这个元素出发,绕数组一圈,能保证累加和一直非负。
代码:
public class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int index = 0;
int total = 0, sum = 0;
if (gas.length == 0 || gas.length != cost.length) {
return -1;
}
for (int i = 0; i < gas.length; i++) {
total += gas[i] - cost[i];
if (sum < 0) {
sum = gas[i] - cost[i];
index = i;
} else {
sum += gas[i] - cost[i];
}
}
return total >= 0 ? index : -1;
}
}
相关文章推荐
- 134. Gas Station
- 134-Gas Station
- LeetCode(134) Gas Station
- 134. Gas Station
- 134. Gas Station
- [Leetcode] #134 Gas Station
- NetBeans 时事通讯(刊号 # 134 - Jan 25, 2011)
- codeforces-div2-134
- 英语练习134 The job interview
- LeetCode _ Gas Station
- LeetCode | Gas Station
- Codeforces Round #134 (Div. 2) A. Mountain Scenery
- Gas Station
- 树形DP求树的重心 --SGU 134
- 树形DP求树的重心 --SGU 134
- SGU 134 Centriod 树的重心
- SGU - 134 Centroid 无根树转有根树 + 树形DP
- Gas Station
- 编写高质量代码改善C#程序的157个建议——建议134:有条件地使用前缀
- SGU 134 树形DP