Keywords: Java

# 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:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

###### Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [- 1, - 1]

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);
}
}
```

Posted by Heatmizer20 on Tue, 05 Nov 2019 13:32:21 -0800