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| 1 | +/* |
| 2 | + * @lc app=leetcode.cn id=704 lang=java |
| 3 | + * |
| 4 | + * [704] 二分查找 |
| 5 | + * |
| 6 | + * https://leetcode-cn.com/problems/binary-search/description/ |
| 7 | + * |
| 8 | + * algorithms |
| 9 | + * Easy (53.71%) |
| 10 | + * Likes: 138 |
| 11 | + * Dislikes: 0 |
| 12 | + * Total Accepted: 52.3K |
| 13 | + * Total Submissions: 96.3K |
| 14 | + * Testcase Example: '[-1,0,3,5,9,12]\n9' |
| 15 | + * |
| 16 | + * 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 |
| 17 | + * target,如果目标值存在返回下标,否则返回 -1。 |
| 18 | + * |
| 19 | + * |
| 20 | + * 示例 1: |
| 21 | + * |
| 22 | + * 输入: nums = [-1,0,3,5,9,12], target = 9 |
| 23 | + * 输出: 4 |
| 24 | + * 解释: 9 出现在 nums 中并且下标为 4 |
| 25 | + * |
| 26 | + * |
| 27 | + * 示例 2: |
| 28 | + * |
| 29 | + * 输入: nums = [-1,0,3,5,9,12], target = 2 |
| 30 | + * 输出: -1 |
| 31 | + * 解释: 2 不存在 nums 中因此返回 -1 |
| 32 | + * |
| 33 | + * |
| 34 | + * |
| 35 | + * |
| 36 | + * 提示: |
| 37 | + * |
| 38 | + * |
| 39 | + * 你可以假设 nums 中的所有元素是不重复的。 |
| 40 | + * n 将在 [1, 10000]之间。 |
| 41 | + * nums 的每个元素都将在 [-9999, 9999]之间。 |
| 42 | + * |
| 43 | + * |
| 44 | + */ |
| 45 | + |
| 46 | +// @lc code=start |
| 47 | +class Solution { |
| 48 | + public int search(int[] nums, int target) { |
| 49 | + int left = 0; |
| 50 | + int right = nums.length - 1; |
| 51 | + while (left <= right) { |
| 52 | + int mid = left + (right - left) / 2; |
| 53 | + if (nums[mid] < target) { |
| 54 | + left = mid + 1; |
| 55 | + } else if (nums[mid] > target) { |
| 56 | + right = mid - 1; |
| 57 | + } else if (nums[mid] == target) { |
| 58 | + return mid; |
| 59 | + } |
| 60 | + } |
| 61 | + return -1; |
| 62 | + } |
| 63 | +} |
| 64 | +// @lc code=end |
| 65 | + |
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