Merge pull request #292 from HCH1212/master

add no.134 answer

leetcode/0134.Gas-Station/Gas.go

@@ -0,0 +1,26 @@
+package leetcode
+
+func canCompleteCircuit(gas []int, cost []int) int {
+	totalGas := 0
+	totalCost := 0
+	currGas := 0
+	start := 0
+
+	for i := 0; i < len(gas); i++ {
+		totalGas += gas[i]
+		totalCost += cost[i]
+		currGas += gas[i] - cost[i]
+
+		if currGas < 0 {
+			start = i + 1
+			currGas = 0
+		}
+	}
+
+	if totalGas < totalCost {
+		return -1
+	}
+
+	return start
+
+}

leetcode/0134.Gas-Station/Gas_test.go

@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question134 struct {
+	para134
+	ans134
+}
+
+// para 是参数
+// one 代表第一个参数
+type para134 struct {
+	one []int
+	two []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans134 struct {
+	one int
+}
+
+func Test_Problem134(t *testing.T) {
+	qs := []question134{
+
+		{
+			para134{[]int{1, 2, 3, 4, 5}, []int{3, 4, 5, 1, 2}},
+			ans134{3},
+		},
+
+		{
+			para134{[]int{2, 3, 4}, []int{3, 4, 3}},
+			ans134{-1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 134------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans134, q.para134
+		fmt.Printf("【input】:%v， %v       【output】:%v\n", p.one, p.two, canCompleteCircuit(p.one, p.two))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0134.Gas-Station/README.md

@@ -0,0 +1,88 @@
+# [134. Gas Stations](https://leetcode.com/problems/gas-station/)
+
+
+## 题目
+
+You are given a list of gas stations on a circular route, where the amount of gas at station i is gas[i], and the cost to travel from station i to its next station (i+1) is cost[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.
+
+Write a function that returns the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique.
+
+**Example 1:**
+
+```
+Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
+Output: 3
+Explanation: Start at station 3 (index 3), you have 4 liters of gas. Now you have a total of =0+4=4 liters of gas.
+Travel to station 4. Now you have 4 - 1 + 5 = 8 liters of gas.
+Travel to station 0. Now you have 8 - 2 + 1 = 7 liters of gas.
+Travel to station 1. Now you have 7 - 3 + 2 = 6 liters of gas.
+Travel to station 2. Now you have 6 - 4 + 3 = 5 liters of gas.
+Travel to station 3 again. This time you will have consumed 5 liters of gas, which is exactly enough for you to return to station 3.
+Therefore, 3 can be the starting index.
+```
+
+**Example 2:**
+
+```
+Input: gas = [2,3,4], cost = [3,4,3]
+Output: -1
+Explanation: You cannot start at station 0 or 1 because there is not enough gas to get you to the next station.
+We start at station 2, we get 4 liters of gas. Now you have a total of =0+4=4 liters of gas.
+Travel to station 0. Now you have 4 - 3 + 2 = 3 liters of gas.
+Travel to station 1. Now you have 3 - 3 + 3 = 3 liters of gas.
+You cannot return to station 2 because it requires 4 liters of gas but you only have 3 liters in your tank.
+Therefore, no matter what, you cannot travel around the ring.
+```
+
+**Constraints:**
+
+- `gas.length == n`
+- `cost.length == n`
+- `1 <= n <= 10^5`
+- `0 <= gas[i], cost[i] <= 10^4`
+
+## 题目大意
+
+在一条环路上有 n 个加油站，其中第 i 个加油站有汽油 gas[i] 升。
+
+你有一辆油箱容量无限的的汽车，从第 i 个加油站开往第 i+1 个加油站需要消耗汽油 cost[i] 升。你从其中的一个加油站出发，开始时油箱为空。
+
+给定两个整数数组 gas 和 cost ，如果你可以按顺序绕环路行驶一周，则返回出发时加油站的编号，否则返回 -1 。如果存在解，则 保证 它是 唯一 的。
+
+## 解题思路
+
+- 初始化总油量（totalGas）和总消耗（totalCost）为0，起始点（start）为0，当前油量（currGas）为0。
+  遍历每个加油站，累加油量和消耗，计算当前油量。
+  如果当前油量小于0，说明从上一个加油站到当前加油站的油量不足以到达下一个加油站，因此将起始点设置为当前加油站的下一个位置，并将当前油量重置为0。
+  遍历结束后，如果总油量小于总消耗，说明无法完成整个旅程，返回-1；否则返回起始点。
+
+## 代码
+
+```go
+package leetcode
+
+func canCompleteCircuit(gas []int, cost []int) int {
+	totalGas := 0
+	totalCost := 0
+	currGas := 0
+	start := 0
+
+	for i := 0; i < len(gas); i++ {
+		totalGas += gas[i]
+		totalCost += cost[i]
+		currGas += gas[i] - cost[i]
+
+		if currGas < 0 {
+			start = i + 1
+			currGas = 0
+		}
+	}
+
+	if totalGas < totalCost {
+		return -1
+	}
+
+	return start
+}
+
+```