subject

Given an ascending array of integers, nums, and a target value, target. Find the start and end position of the given target value in the array.

Your algorithm time complexity must be O(log n) level.

Returns [- 1, - 1] if the target value does not exist in the array.

Example 1:

Example 2:

Thinking of problem solving dichotomy: the time complexity of problem searching must be O(log n), so dichotomy must be adopted for this problem. There are two ways to write:

1. When the middle value of binary search = target, search from the middle value to both sides to get the boundary.
2. When the middle value of binary search = target, continue binary search in the left and right half of the array to find the boundary.

Java problem solving search to both sides

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = {-1, -1};
        if(nums==null || nums.length==0)
            return res;

        int tar = beamSearch(nums, 0, nums.length-1, target);
        if(tar==-1)
            return res;
        int left = tar, right = tar;
        while(left-1>=0 && nums[left-1]==target)
            left--;
        while(right+1<nums.length && nums[right+1]==target)
            right++;
        res[0] = left;
        res[1] = right;
        return res;
    }
    
    public int beamSearch(int[] nums, int left, int right, int target){
        if(left>right)
            return -1;
        int mid = left + (right-left)/2;
        if(nums[mid]==target){
            return mid;
        }else if(nums[mid]<target){
            return beamSearch(nums, mid+1, right, target);
        }else
            return beamSearch(nums, left, mid-1, target);
    }
}

Java problem solving - two binary searches

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] pos = new int[]{Integer.MAX_VALUE,Integer.MIN_VALUE};
        beamSearch(nums,target,0,nums.length-1,pos);
        if(pos[0]==Integer.MAX_VALUE) 
            return new int[]{-1,-1};
        else 
            return pos;
    }

    public void beamSearch(int[] nums,int target,int low,int high,int[] pos){
        if(low>high) return;
        int mid=low+(high-low)/2;
        if(nums[mid]==target){
            if(mid<pos[0]) pos[0]=mid;
            if(mid>pos[1]) pos[1]=mid;
            beamSearch(nums,target,low,mid-1,pos);
            beamSearch(nums,target,mid+1,high,pos);
        }else if(nums[mid]>target)
            beamSearch(nums,target,low,mid-1,pos);
        else
            beamSearch(nums,target,mid+1,high,pos);
    }
}