Gas Station
2014-10-21 20:58
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分解:
1,是否存在?
2,若存在,i 是多少?
-》1,如果一圈的sumAll < 0, 则一定不存在,这个好理解。反之,如果sumAll >= 0, 则一定存在!why?-》反证法,如果每一个i开始绕一圈sum都小于0的话,那么sumAll<0。
2,和 maximum subarray很像啊。这里是维护一个sum > 0的consecutive subarray. 所以当在i时遇到<0了,i之前的就白做功了。 想sum的函数,你必须让它维持在y轴以上,到y轴以下时,则把sum = 0,重新累计。
那么从i到end加起来都>0的话,就一定是问题的解吗?可不可能再加前面的时候又小于0了?这个时候注意note,因为解是unique的,所以只要sumAll>0, i就一定是解。
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
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.
Note:
The solution is guaranteed to be unique.
public class Solution {
public int maxSubArray(int[] A) {
int sum = 0;
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < A.length; i++) {
sum += A[i];
maxSum = Math.max(maxSum, sum);
if (sum < 0) {
sum = 0;
}
}
return maxSum;
}
}
1,是否存在?
2,若存在,i 是多少?
-》1,如果一圈的sumAll < 0, 则一定不存在,这个好理解。反之,如果sumAll >= 0, 则一定存在!why?-》反证法,如果每一个i开始绕一圈sum都小于0的话,那么sumAll<0。
2,和 maximum subarray很像啊。这里是维护一个sum > 0的consecutive subarray. 所以当在i时遇到<0了,i之前的就白做功了。 想sum的函数,你必须让它维持在y轴以上,到y轴以下时,则把sum = 0,重新累计。
那么从i到end加起来都>0的话,就一定是问题的解吗?可不可能再加前面的时候又小于0了?这个时候注意note,因为解是unique的,所以只要sumAll>0, i就一定是解。
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.
Note:
The solution is guaranteed to be unique.
public class Solution {
public int maxSubArray(int[] A) {
int sum = 0;
int maxSum = Integer.MIN_VALUE;
for (int i = 0; i < A.length; i++) {
sum += A[i];
maxSum = Math.max(maxSum, sum);
if (sum < 0) {
sum = 0;
}
}
return maxSum;
}
}
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