Java for LeetCode 034 Search for a Range

標籤：

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm‘s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

解題思路：

看到O(log n) 幾乎可以肯定是二分尋找的思路，題目不是特別難的那種，仔細想想就想出來了，JAVA實現如下：

static public int[] searchRange(int[] nums, int target) {int[] result = new int[2];result[0] = result[1] = -1;int left = 0, right = nums.length - 1;while (left <= right) {if (nums[(left + right) / 2] > target)right = (left + right) / 2 - 1;else if (nums[(left + right) / 2] < target)left = (left + right) / 2 + 1;else {result[0] = result[1] = (left + right) / 2;while (target != nums[left]) {if (target > nums[(result[0] + left) / 2])left = (result[0] + left) / 2 + 1;else {result[0] = (result[0] + left) / 2;left++;}}result[0] = left;while (target != nums[right]) {if (target < nums[(result[1] + right) / 2])right = (result[1] + right) / 2 - 1;else {result[1] = (result[1] + right) / 2;right--;}}result[1] = right;break;}}return result;}

 

Java for LeetCode 034 Search for a Range