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First Position of Target

时间:2016-07-04 06:30:31      阅读:169      评论:0      收藏:0      [点我收藏+]

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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 }

参考请注明出处: cnblogs.com/beiyeqingteng/

First Position of Target

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原文地址:http://www.cnblogs.com/beiyeqingteng/p/5639277.html

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