Search for a Range-优快云博客

本文介绍了一种算法，可以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，该算法的时间复杂度为O(logn)。如果未找到目标值，则返回[-1,-1]。例如，在数组[5,7,7,8,8,10]中查找8，将返回[3,4]。

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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].

public class Solution {
    public int findLeft(int[] nums, int target) {
        int low = 0;
        int high = nums.length - 1;
        while(low <= high) {
            int mid = (low + high) / 2;
            if(nums[mid] < target)
                low = mid + 1;
            else
                high = mid - 1;
        }
        /* need to check the low's index, low will be less than the 0 */
        if(low >= 0 && low < nums.length && nums[low] == target)
            return low;
        else
            return -1;
    }

    public int findRight(int[] nums, int target) {
        int low = 0;
        int high = nums.length - 1;
        while(low <= high) {
            int mid = (low + high) / 2;
            if(nums[mid] <= target)
                low = mid + 1;
            else
                high = mid - 1;
        }
        if(high >= 0 && high < nums.length && nums[high] == target)
            return high;
        else
            return -1;
    }

    public int[] searchRange(int[] nums, int target) {
        int res[] = new int[2];
        res[0] = findLeft(nums, target);
        res[1] = findRight(nums, target);
        return res;
    }
}