Update 11. 旋转数组的最小数字.md

题目链接改为力扣中的对应题目，题目描述和代码按该题的接口调整。

notes/11. 旋转数组的最小数字.md

@@ -2,7 +2,7 @@
 
 ## 题目链接
 
-[牛客网](https://www.nowcoder.com/practice/9f3231a991af4f55b95579b44b7a01ba?tpId=13&tqId=11159&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[Leetcode](https://leetcode.cn/problems/xuan-zhuan-shu-zu-de-zui-xiao-shu-zi-lcof/)
 
 ## 题目描述
 
@@ -24,38 +24,38 @@
 - 否则 [m + 1, h] 区间内的数组是旋转数组，令 l = m + 1。
 
 ```java
-public int minNumberInRotateArray(int[] nums) {
-    if (nums.length == 0)
+public int inventoryManagement(int[] stock) {
+    if (stock.length == 0)
         return 0;
-    int l = 0, h = nums.length - 1;
+    int l = 0, h = stock.length - 1;
     while (l < h) {
         int m = l + (h - l) / 2;
-        if (nums[m] <= nums[h])
+        if (stock[m] <= stock[h])
             h = m;
         else
             l = m + 1;
     }
-    return nums[l];
+    return stock[l];
 }
 ```
 
 如果数组元素允许重复，会出现一个特殊的情况：nums[l] == nums[m] == nums[h]，此时无法确定解在哪个区间，需要切换到顺序查找。例如对于数组 {1,1,1,0,1}，l、m 和 h 指向的数都为 1，此时无法知道最小数字 0 在哪个区间。
 
 ```java
-public int minNumberInRotateArray(int[] nums) {
-    if (nums.length == 0)
+public int inventoryManagement(int[] stock) {
+    if (stock.length == 0)
         return 0;
-    int l = 0, h = nums.length - 1;
+    int l = 0, h = stock.length - 1;
     while (l < h) {
         int m = l + (h - l) / 2;
-        if (nums[l] == nums[m] && nums[m] == nums[h])
-            return minNumber(nums, l, h);
-        else if (nums[m] <= nums[h])
+        if (stock[l] == stock[m] && stock[m] == stock[h])
+            return minNumber(stock, l, h);
+        else if (stock[m] <= stock[h])
             h = m;
         else
             l = m + 1;
     }
-    return nums[l];
+    return stock[l];
 }
 
 private int minNumber(int[] nums, int l, int h) {