LeetCode - Search for a Range

Search for a Range

// Solution 1: 3 binary searches
public int[] searchRange(int[] nums, int target) {
        int[] ret = new int[]{-1, -1};
        if(nums == null || nums.length == 0) return ret;
        int left = 0, right = nums.length - 1;
        int mid = (left + right) / 2;
        while(left < right){
            if(nums[mid] == target){
                ret[0] = mid;
                ret[1] = mid;
                break;
            }else if(nums[mid] < target){
                left = mid + 1;
            }else{
                right = mid - 1;
            }
            mid = (left + right) / 2;
        } 
        if(nums[mid] != target) return ret;
        
        int ll = left, lr = mid - 1;
        int rl = mid + 1, rr = right;
        
        while(ll <= lr){
            int m = (ll + lr) / 2;
            if(nums[m] == target){
                lr = m - 1;
            } else {
                ll = m + 1;
            }
        }
        ret[0] = ll;
        
        while(rl <= rr){
            int m = (rl + rr) / 2;
            if(nums[m] == target){
                rl = m + 1;
            } else {
                rr = m - 1;
            }
        }
        ret[1] = rr;
        
        return ret;
    }
    
    
// Solution 2: 2 binary searches
    public int[] searchRange(int[] A, int target) {
    int[] res = {-1,-1};
    if(A==null || A.length==0)
    {
        return res;
    }
    int ll = 0;
    int lr = A.length-1;
    while(ll<=lr)
    {
        int m = (ll+lr)/2;
        if(A[m]<target)
        {
            ll = m+1;
        }
        else
        {
            lr = m-1;
        }
    }
    int rl = 0;
    int rr = A.length-1;
    while(rl<=rr)
    {
        int m = (rl+rr)/2;
        if(A[m]<=target)
        {
            rl = m+1;
        }
        else
        {
            rr = m-1;
        }
    }
    if(ll<=rr)
    {
        res[0] = ll;
        res[1] = rr;
    }
    return res;
}