添加 0134.加油站.md Scala版本

problems/0134.加油站.md

@@ -471,5 +471,73 @@ int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){
 }
 ```
 
+### Scala
+
+暴力解法:
+
+```scala
+object Solution {
+  def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = {
+    for (i <- cost.indices) {
+      var rest = gas(i) - cost(i)
+      var index = (i + 1) % cost.length // index为i的下一个节点
+      while (rest > 0 && i != index) {
+        rest += (gas(index) - cost(index))
+        index = (index + 1) % cost.length
+      }
+      if (rest >= 0 && index == i) return i
+    }
+    -1
+  }
+}
+```
+
+贪心算法，方法一:
+
+```scala
+object Solution {
+  def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = {
+    var curSum = 0
+    var min = Int.MaxValue
+    for (i <- gas.indices) {
+      var rest = gas(i) - cost(i)
+      curSum += rest
+      min = math.min(min, curSum)
+    }
+    if (curSum < 0) return -1 // 情况1: gas的总和小于cost的总和，不可能到达终点
+    if (min >= 0) return 0 // 情况2: 最小值>=0，从0号出发可以直接到达
+    // 情况3: min为负值，从后向前看，能把min填平的节点就是出发节点
+    for (i <- gas.length - 1 to 0 by -1) {
+      var rest = gas(i) - cost(i)
+      min += rest
+      if (min >= 0) return i
+    }
+    -1
+  }
+}
+```
+
+贪心算法，方法二:
+
+```scala
+object Solution {
+  def canCompleteCircuit(gas: Array[Int], cost: Array[Int]): Int = {
+    var curSum = 0
+    var totalSum = 0
+    var start = 0
+    for (i <- gas.indices) {
+      curSum += (gas(i) - cost(i))
+      totalSum += (gas(i) - cost(i))
+      if (curSum < 0) {
+        start = i + 1 // 起始位置更新
+        curSum = 0  // curSum从0开始
+      }
+    }
+    if (totalSum < 0) return -1 // 说明怎么走不可能跑一圈
+    start
+  }
+}
+```
+
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