From df5a6c62d50d0473745ce3406ceb4e04864fded1 Mon Sep 17 00:00:00 2001
From: vanyongqi <46806467+vanyongqi@users.noreply.github.com>
Date: Tue, 20 Sep 2022 18:17:56 +0800
Subject: [PATCH] =?UTF-8?q?Update=200704.=E4=BA=8C=E5=88=86=E6=9F=A5?=
 =?UTF-8?q?=E6=89=BE.md?=
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增加了二分查找C语言的左闭右开区间的版本，已通过leeitcode的验证
---
 problems/0704.二分查找.md | 25 ++++++++++++++++++++++++-
 1 file changed, 24 insertions(+), 1 deletion(-)

diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md
index 53f5331a..26d821bd 100644
--- a/problems/0704.二分查找.md
+++ b/problems/0704.二分查找.md
@@ -515,7 +515,7 @@ impl Solution {
 ```
 
 **C:**
-```c
+```c(版本一)
 int search(int* nums, int numsSize, int target){
     int left = 0;
     int right = numsSize-1;
@@ -541,6 +541,29 @@ int search(int* nums, int numsSize, int target){
     return -1;
 }
 ```
+```
+C (版本二)
+int search(int* nums, int numsSize, int target){
+    int length = numsSize;
+    int left = 0;
+    int right = length;//定义target在左闭右开的区间里，即：[left, right)
+    int middle = 0;
+    while(left < right){// left == right时，区间[left, right)属于空集，所以用 < 避免该情况
+        int middle = left + (right - left) / 2;
+        if(nums[middle] < target){
+            //target位于(middle , right) 中为保证集合区间的左闭右开性，可等价为[middle + 1,right)
+            left = middle + 1;
+        }else if(nums[middle] > target){
+            //target位于[left, middle)中
+            right = middle ;
+        }else{// nums[middle] == target ，找到目标值target
+            return middle;
+        }
+    }
+    //未找到目标值，返回-1
+    return -1;
+}
+```
 
 **PHP:**
 ```php