Search for a Range

Description:

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].

Ex:

[5, 7, 7, 8, 8, 10] and target value 8—–[3, 4]
[5, 7, 7, 8, 10] and target value 8—–[3 , 3]
[5, 7, 7, 10] and target value 8—-[-1, -1]

问题描述：

搜索范围，给一串递增的数组和目标元素，返回目标元素在数组中出现的起始位置索引，如果目标元素不存在数组中，返回[-1 -1]

解法一：

思路：

总体上看是两次二分查找，时间复杂度为O(logn)

Code:

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] result = new int[2];
        result[1] = findFirst(nums,target);
        result[0]= findLast(nums,target);
        return result;
    }

    private int findFirst(int[] nums, int target){
        int idx = -1;
        int start = 0;
        int end = nums.length -1;

        while(start <= end){
            int mid = start + (end - start)/2;
            if(nums[mid] > target){
                end = mid - 1;
            }
            else{
                start = mid + 1;
            }
            if(nums[mid] == target) idx = mid;
        }
        return idx;
    }

    private int findLast(int[] nums,int target){
        int idx = -1;
        int start = 0;
        int end = nums.length -1;

        while(start <= end){
            int mid = start + (end - start)/2;
            if(nums[mid] < target){
                start = mid + 1;
            }
            else{
                end = mid -1;
            }
            if(nums[mid] == target) idx = mid;
        }
        return idx;
    }
}

备注

这里的两个函数太相似了，而且逻辑有点昏。。。。