L34 Find First and Last Position of Element in Sorted Array

Problem:

Given an array of integers nums 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].

example:

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

Solution

Binary Search:

There are two parts for this question: search left and search right.

for the left part, since we need to find the most left index, so when nums[mid] < target, left = mid; the point always stay at most left.

TC: log n; SC:O1

public int[] searchRange(int[] nums, int target){
    if(nums == null || nums.length == 0) return new int{-1, -1};
    int[] res = new int[2];
    res[0] = searchLeft(nums, target);
    res[1] = searchRight(nums, target);
    return res;
}
private int searchLeft(int[]nums, int target){
    int left = 0, right = nums.length-1;
    while(left+1 < right){
        int mid = left + (right - left) / 2;
        if(nums[mid] >= target){
            right = mid;
        }else{
            left = mid;
        }
    }
    if(nums[left] == target) return left;
    return nums[right] == target? right : -1;
}
private int searchRight(int[]nums, int target){
    int left = 0, right = nums.length - 1;
    while(left + 1 < right){
        int mid = left + (right - left) / 2;
        if(mid <= target){
            left = mid;
        }else{
            right = mid;
        }
    }
    if(nums[right] == target) return right;
    return nums[left] == target ? left : -1;
}