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DAY 7
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Software Development

LeetCode小試身手系列 第 7

【Day 7】#1184 - Distance Between Bus Stops

A bus has n stops numbered from 0 to n - 1 that form a circle. We know the distance between all pairs of neighboring stops where distance[i] is the distance between the stops number i and (i + 1) % n.

The bus goes along both directions i.e. clockwise and counterclockwise.

Return the shortest distance between the given start and destination stops.

Example 1:

Input: distance = [1,2,3,4], start = 0, destination = 1
Output: 1
Explanation: Distance between 0 and 1 is 1 or 9, minimum is 1.

Example 2:

Input: distance = [1,2,3,4], start = 0, destination = 2
Output: 3
Explanation: Distance between 0 and 2 is 3 or 7, minimum is 3.

Example 3:

Input: distance = [1,2,3,4], start = 0, destination = 3
Output: 4
Explanation: Distance between 0 and 3 is 6 or 4, minimum is 4.

Constraints:

  • 1 <= n <= 10^4
  • distance.length == n
  • 0 <= start, destination < n
  • 0 <= distance[i] <= 10^4

解析

此題要求起點與終點的最短路徑,可分別以順時針及逆時針兩種情形來求解。

解法

class Solution {
	public int distanceBetweenBusStops(int[] distance, int start, int destination) {
		if (start == destination) {
			return 0;
		}
		int clockwiseSum = 0, counterclockwiseSum = 0, begin = start, end = destination;

		// Calculate the distance clockwise
		if (destination % distance.length < start % distance.length) {
			// To know how many stops between start and destination
			end += distance.length;
		}
		for (int i = begin; i < end; i++) {
			int busStop = i % distance.length;
			clockwiseSum += distance[busStop];
		}

		// Calculate the distance counterclockwise
		begin = start;
		end = destination;
		if (destination % distance.length > start % distance.length) {
			// To know how many stops between start and destination
			begin += distance.length;
		}
		for (int i = begin - 1; i >= end; i--) {
			int busStop = i % distance.length;
			counterclockwiseSum += distance[busStop];
		}

		// Compare both distances to find out the shortest distance
		if (clockwiseSum < counterclockwiseSum) {
			return clockwiseSum;
		} else {
			return counterclockwiseSum;
		}
	}
}

心得

此題難度是Contest的題目,難度為Easy。


希望透過記錄解題的過程,可以對於資料結構及演算法等有更深一層的想法。
如有需訂正的地方歡迎告知,若有更好的解法也歡迎留言,謝謝。


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【Day 8】#136 - Single Number
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