From 60cff832db8e802eaf94850fd8b303041d6586e9 Mon Sep 17 00:00:00 2001
From: whai <lkwhai@163.com>
Date: Sat, 2 Dec 2023 20:56:26 +0800
Subject: [PATCH] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E4=BA=8C=E5=88=86=E6=9F=A5?=
 =?UTF-8?q?=E6=89=BE704?=
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Content-Type: text/plain; charset=UTF-8
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---
 pom.xml                                       |   7 +
 .../whaifree/leetCode/easy/LeetCode704.java   | 132 ++++++++++++++++++
 2 files changed, 139 insertions(+)
 create mode 100644 src/main/java/cn/whaifree/leetCode/easy/LeetCode704.java

diff --git a/pom.xml b/pom.xml
index c4251f0..ac263f8 100644
--- a/pom.xml
+++ b/pom.xml
@@ -17,6 +17,13 @@
 <dependencies>
 
 
+
+
+	<dependency>
+		<groupId>junit</groupId>
+		<artifactId>junit</artifactId>
+		<version>4.12</version>
+	</dependency>
 </dependencies>
 
 </project>
diff --git a/src/main/java/cn/whaifree/leetCode/easy/LeetCode704.java b/src/main/java/cn/whaifree/leetCode/easy/LeetCode704.java
new file mode 100644
index 0000000..5442e31
--- /dev/null
+++ b/src/main/java/cn/whaifree/leetCode/easy/LeetCode704.java
@@ -0,0 +1,132 @@
+package cn.whaifree.leetCode.easy;
+
+import org.junit.Assert;
+import org.junit.Test;
+
+/**
+ *
+ * 二分查找
+ *
+ *
+ 给定一个 n 个元素有序的（升序）整型数组 nums 和一个目标值 target  ，写一个函数搜索 nums 中的 target，如果目标值存在返回下标，否则返回 -1。
+ 示例 1:
+
+ 输入: nums = [-1,0,3,5,9,12], target = 9
+ 输出: 4
+ 解释: 9 出现在 nums 中并且下标为 4
+ 示例 2:
+
+ 输入: nums = [-1,0,3,5,9,12], target = 2
+ 输出: -1
+ 解释: 2 不存在 nums 中因此返回 -1
+
+
+ 提示：
+
+ 你可以假设 nums 中的所有元素是不重复的。
+ n 将在 [1, 10000]之间。
+ nums 的每个元素都将在 [-9999, 9999]之间。
+ */
+public class LeetCode704 {
+
+
+    /**
+     * 二分查找 [left,right]
+     * @param nums
+     * @param target
+     * @return
+     */
+    public static int search(int[] nums, int target) {
+        if (nums.length == 0) {
+            return -1;
+        }
+        // 初始化指针
+        int left = 0;
+        int right = nums.length - 1;
+        int index = 0;
+        while (left < right) {
+            // 计算中间索引
+            index = (right + left) / 2;
+            if (nums[index] >= target) {
+                right = index;
+            } else{
+                left = index + 1;
+            }
+        }
+        if (nums[(right + left) / 2] == target) {
+            return (right + left) / 2;
+        }
+
+
+        return -1;
+
+    }
+
+    public int search1(int[] nums, int target) {
+        int left = 0; // 左指针，初始指向数组最左边的位置
+        int right = nums.length - 1; // 右指针，初始指向数组最右边的位置
+        while (left <= right) { // 当左指针小于等于右指针时，循环执行
+            int mid = left + (right - left) / 2; // 中间指针，计算左指针和右指针的中间位置
+            if (nums[mid] == target) { // 如果中间位置的数值等于目标数值
+                return mid; // 返回中间位置的索引
+            } else if (nums[mid] < target) { // 如果中间位置的数值小于目标数值
+                left = mid + 1; // 将左指针移动到中间位置的右边
+            } else { // 如果中间位置的数值大于目标数值
+                right = mid - 1; // 将右指针移动到中间位置的左边
+            }
+        }
+        return -1; // 没有找到目标数值，返回-1
+    }
+
+
+    /**
+     * 二分查找 [left,right)
+     * @param nums
+     * @param target
+     * @return
+     */
+    public int search2(int[] nums, int target) {
+        int left = 0;
+        int right = nums.length;
+        while (left < right) {
+            int middle = (left + right) / 2;
+            if (nums[left] == target) {
+                return left;
+            } else if (nums[middle] >= target) {
+                right = middle;
+            } else {
+                left = middle+1;
+            }
+        }
+
+        return -1;
+
+    }
+
+
+
+
+
+    @Test
+    public void testSearch1() {
+        LeetCode704 solution = new LeetCode704();
+
+        int[] nums = {1};
+        int target = 5;
+        Assert.assertEquals(-1, solution.search2(nums, target));
+
+        nums = new int[]{10, 20, 30, 40};
+        target = 30;
+        Assert.assertEquals(2, solution.search1(nums, target));
+
+        nums = new int[]{1, 2, 3};
+        target = 4;
+        Assert.assertEquals(-1, solution.search1(nums, target));
+
+
+        nums = new int[]{1};
+        target = 1;
+        Assert.assertEquals(0, solution.search1(nums, target));
+
+    }
+}