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feat: add solutions to lc problem: No.0134
No.0134.Gas Station
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‎solution/0100-0199/0134.Gas Station/README.md

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Original file line numberDiff line numberDiff line change
@@ -56,22 +56,110 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：从任意起点开始遍历**
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我们用 $i$, $j$ 分别标记起点和终点，用 $s$ 表示当前剩余汽油，而 $cnt$ 表示当前行驶过的加油站数量。初始时，我们将起点设在最后一个位置，即 $i=n-1$。
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开始行驶，移动 $j$。若发现当前剩余汽油小于 $0$，说明当前 $i$ 作为起点不符合要求，我们将起点 $i$ 循环左移，并且更新剩余汽油，直至剩余汽油是非负数。
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当行驶过的加油站数量达到 $n$ 时，结束。判断此时的剩余汽油是否非负，是则返回当前的 $i$ 作为答案；否则返回 $-1$，表示无解。
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时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 表示加油站的数量。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
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n = len(gas)
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i = j = n - 1
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cnt = s = 0
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while cnt < n:
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s += gas[j] - cost[j]
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cnt += 1
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j = (j + 1) % n
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while s < 0 and cnt < n:
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i -= 1
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s += gas[i] - cost[i]
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cnt += 1
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return -1 if s < 0 else i
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int canCompleteCircuit(int[] gas, int[] cost) {
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int n = gas.length;
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int i = n - 1, j = n - 1;
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int cnt = 0, s = 0;
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while (cnt < n) {
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s += gas[j] - cost[j];
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++cnt;
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j = (j + 1) % n;
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while (s < 0 && cnt < n) {
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--i;
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s += gas[i] - cost[i];
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++cnt;
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}
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}
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return s < 0 ? -1 : i;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
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int n = gas.size();
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int i = n - 1, j = n - 1;
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int cnt = 0, s = 0;
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while (cnt < n) {
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s += gas[j] - cost[j];
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++cnt;
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j = (j + 1) % n;
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while (s < 0 && cnt < n) {
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--i;
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s += gas[i] - cost[i];
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++cnt;
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}
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}
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return s < 0 ? -1 : i;
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}
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};
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```
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### **Go**
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```go
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func canCompleteCircuit(gas []int, cost []int) int {
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n := len(gas)
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i, j := n - 1, n - 1
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cnt, s := 0, 0
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for cnt < n {
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s += gas[j] - cost[j]
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cnt++
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j = (j + 1) % n
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for s < 0 && cnt < n {
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i--
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s += gas[i] - cost[i]
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cnt++
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}
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}
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if s < 0 {
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return -1
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}
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return i
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}
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```
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### **...**

‎solution/0100-0199/0134.Gas Station/README_EN.md

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@@ -56,13 +56,91 @@ Therefore, you can&#39;t travel around the circuit once no matter where you star
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### **Python3**
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```python
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class Solution:
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def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
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n = len(gas)
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i = j = n - 1
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cnt = s = 0
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while cnt < n:
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s += gas[j] - cost[j]
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cnt += 1
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j = (j + 1) % n
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while s < 0 and cnt < n:
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i -= 1
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s += gas[i] - cost[i]
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cnt += 1
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return -1 if s < 0 else i
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```
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### **Java**
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```java
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class Solution {
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public int canCompleteCircuit(int[] gas, int[] cost) {
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int n = gas.length;
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int i = n - 1, j = n - 1;
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int cnt = 0, s = 0;
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while (cnt < n) {
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s += gas[j] - cost[j];
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++cnt;
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j = (j + 1) % n;
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while (s < 0 && cnt < n) {
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--i;
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s += gas[i] - cost[i];
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++cnt;
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}
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}
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return s < 0 ? -1 : i;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
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int n = gas.size();
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int i = n - 1, j = n - 1;
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int cnt = 0, s = 0;
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while (cnt < n) {
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s += gas[j] - cost[j];
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++cnt;
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j = (j + 1) % n;
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while (s < 0 && cnt < n) {
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--i;
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s += gas[i] - cost[i];
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++cnt;
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}
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}
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return s < 0 ? -1 : i;
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}
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};
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```
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### **Go**
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```go
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func canCompleteCircuit(gas []int, cost []int) int {
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n := len(gas)
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i, j := n - 1, n - 1
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cnt, s := 0, 0
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for cnt < n {
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s += gas[j] - cost[j]
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cnt++
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j = (j + 1) % n
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for s < 0 && cnt < n {
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i--
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s += gas[i] - cost[i]
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cnt++
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}
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}
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if s < 0 {
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return -1
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}
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return i
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}
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```
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### **...**

‎solution/0100-0199/0134.Gas Station/Solution.cpp

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class Solution {
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public:
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int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
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int n = gas.size();
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int i = n - 1, j = n - 1;
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int cnt = 0, s = 0;
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while (cnt < n) {
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s += gas[j] - cost[j];
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++cnt;
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j = (j + 1) % n;
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while (s < 0 && cnt < n) {
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--i;
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s += gas[i] - cost[i];
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++cnt;
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}
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}
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return s < 0 ? -1 : i;
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}
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};

‎solution/0100-0199/0134.Gas Station/Solution.go

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func canCompleteCircuit(gas []int, cost []int) int {
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n := len(gas)
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i, j := n - 1, n - 1
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cnt, s := 0, 0
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for cnt < n {
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s += gas[j] - cost[j]
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cnt++
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j = (j + 1) % n
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for s < 0 && cnt < n {
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i--
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s += gas[i] - cost[i]
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cnt++
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}
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}
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if s < 0 {
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return -1
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}
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return i
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}

‎solution/0100-0199/0134.Gas Station/Solution.java

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class Solution {
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public int canCompleteCircuit(int[] gas, int[] cost) {
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if(gas.length!=cost.length) return -1;
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int sum=0,total=0,index=0;
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for(int i=0;i<gas.length;i++){
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int sy = gas[i] - cost[i];
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sum+= sy;
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total+= sy;
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if(sum<0){
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index=i+1;
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sum=0;
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int n = gas.length;
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int i = n - 1, j = n - 1;
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int cnt = 0, s = 0;
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while (cnt < n) {
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s += gas[j] - cost[j];
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++cnt;
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j = (j + 1) % n;
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while (s < 0 && cnt < n) {
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--i;
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s += gas[i] - cost[i];
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++cnt;
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}
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}
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if(total<0) return -1;
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return index;
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return s < 0 ? -1 : i;
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}
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}

‎solution/0100-0199/0134.Gas Station/Solution.py

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class Solution:
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def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
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n = len(gas)
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i = j = n - 1
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cnt = s = 0
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while cnt < n:
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s += gas[j] - cost[j]
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cnt += 1
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j = (j + 1) % n
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while s < 0 and cnt < n:
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i -= 1
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s += gas[i] - cost[i]
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cnt += 1
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return -1 if s < 0 else i

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