First Position of Target-优快云博客

本文介绍了一种在已排序数组中使用二分查找法找到指定目标值首个出现位置的算法实现，时间复杂度为O(logn)。通过具体代码示例展示了如何处理重复元素的情况，确保返回的是目标值的第一个索引。

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For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.

If the target number does not exist in the array, return -1.

Example

If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

分析：
很明显的用binary search, 但是因为要找第一个，所以，我们需要判断一下。

 1 class Solution {
 2     /**
 3      * @param nums: The integer array.
 4      * @param target: Target to find.
 5      * @return: The first position of target. Position starts from 0.
 6      */
 7     public int binarySearch(int[] nums, int target) {
 8         if (nums == null || nums.length == 0) return -1;
 9         
10         int start = 0;
11         int end = nums.length - 1;
12         
13         while (start <= end) {
14             int mid = start + (end - start) / 2;
15             if (nums[mid] == target) {
16                 if (mid != 0 && nums[mid - 1] == target) {
17                     end = mid - 1;
18                 } else {
19                     return mid;
20                 }
21             } else if (nums[mid] < target) {
22                 start = mid + 1;
23             } else {
24                 end = mid - 1;
25             }
26         }
27         return -1;
28     }
29 }