From d19cd4f07244c0d003d39d3f9225655e7bb3bdcb Mon Sep 17 00:00:00 2001
From: whai <lkwhai@163.com>
Date: Wed, 6 Dec 2023 08:18:12 +0800
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---
 .../whaifree/leetCode/middle/LeetCode34.java  | 109 ++++++++++++++++++
 1 file changed, 109 insertions(+)
 create mode 100644 src/main/java/cn/whaifree/leetCode/middle/LeetCode34.java

diff --git a/src/main/java/cn/whaifree/leetCode/middle/LeetCode34.java b/src/main/java/cn/whaifree/leetCode/middle/LeetCode34.java
new file mode 100644
index 0000000..54dbe6c
--- /dev/null
+++ b/src/main/java/cn/whaifree/leetCode/middle/LeetCode34.java
@@ -0,0 +1,109 @@
+package cn.whaifree.leetCode.middle;
+
+import org.junit.Test;
+
+/**
+ * 34. 在排序数组中查找元素的第一个和最后一个位置
+ * 中等
+ * 给你一个按照非递减顺序排列的整数数组 nums，和一个目标值 target。请你找出给定目标值在数组中的开始位置和结束位置。
+ *
+ * 如果数组中不存在目标值 target，返回 [-1, -1]。
+ *
+ * 你必须设计并实现时间复杂度为 O(log n) 的算法解决此问题。
+ * 示例 1：
+ *
+ * 输入：nums = [5,7,7,8,8,10], target = 8
+ * 输出：[3,4]
+ * 示例 2：
+ *
+ * 输入：nums = [5,7,7,8,8,10], target = 6
+ * 输出：[-1,-1]
+ * 示例 3：
+ *
+ * 输入：nums = [], target = 0
+ * 输出：[-1,-1]
+ *
+ *
+ * 提示：
+ *
+ * 0 <= nums.length <= 105
+ * -109 <= nums[i] <= 109
+ * nums 是一个非递减数组
+ * -109 <= target <= 109
+ */
+public class LeetCode34 {
+    /**
+     * 非递减顺序排列的整数数组
+     * [0,length]
+     * [5,8,8,8,8,10]
+     * @param nums
+     * @param target
+     * @return
+     */
+    public int[] searchRange(int[] nums, int target) {
+        int[] ans = new int[2];
+        // 三头
+        int left = 0;
+        int right = nums.length - 1;
+        while (left <= right) {
+            int middle = (left + right) / 2;
+            if (nums[middle] == target) {
+                // 在[left，middle) , (middle,right]区间里找到第一个和最后一个target
+
+//                break;
+                // 找到了一个
+                if (nums[left] == target && nums[right] == target) {
+                    ans[0] = left;
+                    ans[1] = right;
+                    return ans;
+                } else if (nums[left] == target) {
+                    right--;
+                } else if (nums[right] == target) {
+                    left++;
+                } else {
+                    right--;
+                    left++;
+                }
+            } else if (nums[middle] < target) {
+                left = middle + 1;
+            } else {
+                right = middle - 1;
+            }
+        }
+        ans[0] = -1;
+        ans[1] = -1;
+        return ans;
+    }
+    public int[] searchRange2(int[] nums, int target) {
+        int leftIdx = binarySearch(nums, target, true);
+        int rightIdx = binarySearch(nums, target, false) - 1;
+        if (leftIdx <= rightIdx && rightIdx < nums.length && nums[leftIdx] == target && nums[rightIdx] == target) {
+            return new int[]{leftIdx, rightIdx};
+        }
+        return new int[]{-1, -1};
+    }
+
+    public int binarySearch(int[] nums, int target, boolean lower) {
+        int left = 0, right = nums.length - 1, ans = nums.length;
+        while (left <= right) {
+            int mid = (left + right) / 2;
+            if (nums[mid] > target || (lower && nums[mid] >= target)) {
+                right = mid - 1;
+                ans = mid;
+            } else {
+                left = mid + 1;
+            }
+        }
+        return ans;
+    }
+
+
+
+    @Test
+    public void test() {
+        for (int i : searchRange2(new int[]{5,8,8,8,8,10}, 8)) {
+            System.out.println(i);
+        }
+    }
+
+}