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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
.
If the array is [1, 2, 3, 3, 4, 5, 10]
, for given target 3
, return 2
.
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.length == 0) { return -1; } int start = 0, 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) { start = mid; }else { end = mid; } } if (nums[start] == target) { return start; } if (nums[end] == target) { return end; } return -1; } }
[lintcode 14] First Position of Target
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原文地址:http://www.cnblogs.com/iwangzheng/p/5747514.html