solve 33.搜索旋转排序数组

Author: andavid

zh/33.搜索旋转排序数组.java

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+/*
+ * @lc app=leetcode.cn id=33 lang=java
+ *
+ * [33] 搜索旋转排序数组
+ *
+ * https://leetcode-cn.com/problems/search-in-rotated-sorted-array/description/
+ *
+ * algorithms
+ * Medium (37.95%)
+ * Likes:    821
+ * Dislikes: 0
+ * Total Accepted:    140.8K
+ * Total Submissions: 368.2K
+ * Testcase Example:  '[4,5,6,7,0,1,2]\n0'
+ *
+ * 假设按照升序排序的数组在预先未知的某个点上进行了旋转。
+ *
+ * ( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。
+ *
+ * 搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。
+ *
+ * 你可以假设数组中不存在重复的元素。
+ *
+ * 你的算法时间复杂度必须是 O(log n) 级别。
+ *
+ * 示例 1:
+ *
+ * 输入: nums = [4,5,6,7,0,1,2], target = 0
+ * 输出: 4
+ *
+ *
+ * 示例 2:
+ *
+ * 输入: nums = [4,5,6,7,0,1,2], target = 3
+ * 输出: -1
+ *
+ */
+
+// @lc code=start
+class Solution {
+  public int search(int[] nums, int target) {
+    int left = 0;
+    int right = nums.length - 1;
+
+    while (left <= right) {
+      int mid = left + (right - left) / 2;
+      if (nums[mid] == target) {
+        return mid;
+      } else if (nums[left] <= nums[mid]) {
+        if (nums[left] <= target && target < nums[mid]) {
+          right = mid - 1;
+        } else {
+          left = mid + 1;
+        }
+      } else {
+        if (nums[mid] < target && target <= nums[right]) {
+          left = mid + 1;
+        } else {
+          right = mid - 1;
+        }
+      }
+    }
+
+    return -1;
+  }
+}
+// @lc code=end
+