关联LeetCode题号134

本题特点

• 贪心
• 局部最优解-部分差值 如果小于0（消耗大于油站油量） 就从下一个加油站开始，因为如果中间有小于0的情况 当前站就不可能是始发站，
• 整体最优解-整体差值 如果小于0 ，那么就是不能有始发站

本题思路

class Solution:def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:curSum = 0totalSum = 0start = 0for i in range(len(gas)):curSum += gas[i] - cost[i]totalSum += gas[i] - cost[i]if curSum < 0:start = i + 1curSum = 0if totalSum < 0:return -1return start

java写法

package leetcode;import org.junit.jupiter.api.Test;/*** File Description: gasStation_134* Author: Your Name* Date: 2024/12/24*/
public class gasStation_134 {public int canCompleteCircuit(int[] gas, int[] cost) {int curSum = 0;int totalSum = 0;int start = 0;for (int i = 0; i < gas.length; i++){curSum += gas[i] - cost[i];totalSum += gas[i] - cost[i];if(curSum < 0){start = i + 1;curSum = 0;}}if (totalSum < 0){return -1;}return start;}@Testpublic void TestCanCompleteCircuit() {int[] gas = {2,3,4};int[] cost = {3,4,3};int start = canCompleteCircuit(gas, cost);System.out.println(start);}}