From d327806115b66777e8e19ee27af10d4491b7a609 Mon Sep 17 00:00:00 2001
From: ArthurP <gpan20020205@gmail.com>
Date: Sun, 29 Aug 2021 10:37:39 +0800
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---
 problems/0134.加油站.md | 7 ++++++-
 1 file changed, 6 insertions(+), 1 deletion(-)

diff --git a/problems/0134.加油站.md b/problems/0134.加油站.md
index ff54fbc6..ff3bc8b2 100644
--- a/problems/0134.加油站.md
+++ b/problems/0134.加油站.md
@@ -289,21 +289,26 @@ int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){
     int curSum = 0;
     int i;
     int min = INT_MAX;
+    //遍历整个数组。计算出每站的用油差。并将其与最小累加量比较
     for(i = 0; i < gasSize; i++) {
         int diff = gas[i] - cost[i];
         curSum += diff;
-        if(min > curSum)
+        if(curSum < min)
             min = curSum;
     }
+    //若汽油总数为负数，代表无法跑完一环。返回-1
     if(curSum < 0)
         return -1;
+    //若min大于等于0，说明每一天加油量比用油量多。因此从0出发即可
     if(min >= 0)
         return 0;
+    //若累加最小值为负，则找到一个非零元素（加油量大于出油量）出发。返回坐标
     for(i = gasSize - 1; i >= 0; i--) {
         min+=(gas[i]-cost[i]);
         if(min >= 0)
             return i;
     }
+    //逻辑上不会返回这个0
     return 0;
 }
 ```