LintCode 14. First Position of Target 原创Java参考解答
问题描述
http://www.lintcode.com/en/problem/first-position-of-target/
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
.
解题思路
- 先检查数组nums是否为null或者为空数组,若是则直接返回-1。
- 用二分法在数组中查找target值的位置。若查到了,不直接返回该target位置,而是继续以该位置开始的前面部分继续进行二分查找。以此来查找target在数组中第一次出现的位置。
参考代码
class Solution { /** * @param nums: The integer array. * @param target: Target to find. * @return: The first position of target. Position starts from 0. */ public int binarySearch(int[] nums, int target) { if (nums == null || nums.length == 0) { return -1; } int start = 0; int end = nums.length - 1; while (start + 1 < end) { int mid = start + (end - start) / 2; if (nums[mid] == target) { end = mid; } else if (nums[mid] > target) { end = mid; } else { start = mid; } } if (nums[start] == target) { return start; } else if (nums[end] == target) { return end; } return -1; } }