From fa934cf225cc37f080d3742211a983febf9e4f07 Mon Sep 17 00:00:00 2001
From: Sainnhe Park <i@sainnhe.dev>
Date: Tue, 7 Feb 2023 11:44:09 +0800
Subject: [PATCH] s0134

---
 include/s0134_gas_station.hpp | 13 +++++++++++++
 src/s0134_gas_station.cpp     | 27 +++++++++++++++++++++++++++
 2 files changed, 40 insertions(+)
 create mode 100644 include/s0134_gas_station.hpp
 create mode 100644 src/s0134_gas_station.cpp

diff --git a/include/s0134_gas_station.hpp b/include/s0134_gas_station.hpp
new file mode 100644
index 0000000..788fab7
--- /dev/null
+++ b/include/s0134_gas_station.hpp
@@ -0,0 +1,13 @@
+#ifndef S0134_GAS_STATION_HPP
+#define S0134_GAS_STATION_HPP
+
+#include <vector>
+
+using namespace std;
+
+class S0134 {
+ public:
+  int canCompleteCircuit(vector<int>& gas, vector<int>& cost);
+};
+
+#endif
diff --git a/src/s0134_gas_station.cpp b/src/s0134_gas_station.cpp
new file mode 100644
index 0000000..3c1e948
--- /dev/null
+++ b/src/s0134_gas_station.cpp
@@ -0,0 +1,27 @@
+#include "s0134_gas_station.hpp"
+
+// 我们假设 rest[i] = gas[i] - cost[i] 等于当前加油站所能积累的油量
+// 情况一：
+// 如果 totalSum = rest[0] + rest[1] + ... + rest[n-1] < 0 则说明所有加油站的油加起来都不够跑一圈的
+// 情况二：
+// 我们假设 totalSum >= 0 也就是所有加油站的油加起来能跑一圈
+// 当前加油站索引为 i ，起始节点为 startIndex
+// 那么如果 curSum = rest[startIndex] + rest[startIndex+1] + ... + rest[i] < 0 则说明剩下的节点加起来一定大于零
+// 也就是 i 之前出现了多少负数，之后就应该出现多少正数
+// 这时把 startIndex 设为 i+1 ，curSum = 0 继续迭代
+int S0134::canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
+  int curSum{0};
+  int totalSum{0};
+  int startIndex{0};
+  int len = gas.size();
+  for (int i{0}; i < len; ++i) {
+    curSum += gas[i] - cost[i];
+    totalSum += gas[i] - cost[i];
+    if (curSum < 0) {
+      startIndex = i + 1;
+      curSum = 0;
+    }
+  }
+  if (totalSum < 0) return -1;
+  return startIndex;
+}