34. Search for a Range

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本文介绍了一种算法，能够在已排序的整数数组中找到给定目标值的起始和结束位置，采用二分查找法分别定位左右边界，确保了算法的时间复杂度为O(log n)。

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Given an array of integers sorted in ascending order, 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].

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Solution:

Tips:

find right and left index respectively leveraging Binary search。

Java Code:

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;
        int[] result = {-1, -1};
        
        if (null == nums || nums.length < 1) {
            return result;
        }
        
        // find the left index
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (target == nums[mid]) {
                if (0 == mid || target > nums[mid - 1]) {
                    result[0] = mid;
                    break;
                }
                right = mid - 1;
            } else if (nums[mid] > target) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }

        left = result[0] < 0 ? 0 : result[0];
        right = nums.length - 1;
        // find the right index
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (target == nums[mid]) {
                if (nums.length - 1 == mid || target < nums[mid + 1]) {
                    result[1] = mid;
                    break;
                }
                left = mid + 1;
            } else if (nums[mid] > target) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        
        return result;
    }
}