[Leetcode + Lintcode] 34. Search for a Range

Given a sorted array of integers, 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].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

用binary search轻松解决，search for a range 等同于search for the first/last appearance，只需对常规的binary search 稍作改动，不要一看到等于target就返回index,而是应该根据找first/last来判断是否还要继续做binary search (ex.寻找first appearance时，如果nums[mid] >= target都应该继续binary search,因为找到的index并不一定是first appearance，应该进一步向前寻找直到跳出循环，还应优先判断nums[start]==target是否成立来决定first index)

1. java

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        if(nums == null){
            return new int[]{-1,-1};
        }
        int start = 0;
        int end = nums.length - 1;
        while(start + 1 < end){
            int mid = start + (end - start)/2;
            if(nums[mid] < target){
                start = mid;
            }
            else if(nums[mid] >= target){
                end = mid;
            }
        }
        int[] result = new int[2];
        if(nums[start] == target){
            result[0] = start;
        }
        else if(nums[end] == target){
            result[0] = end;
        }
        else {
            return new int[]{-1,-1};
        }
        start = 0;
        end = nums.length - 1;
        while(start + 1 < end){
            int mid = start + (end - start)/2;
            if(nums[mid] <= target){
                start = mid;
            }
            else if(nums[mid] > target){
                end = mid;
            }
        }
        if(nums[end] == target){
            result[1] = end;
            return result;
        }
        else if(nums[start] == target){
            result[1] = start;
            return result;
        }
        return new int[]{-1,-1};
    }
}

2.Python

class Solution:
    """
    @param A : a list of integers
    @param target : an integer to be searched
    @return : a list of length 2, [index1, index2]
    """
    def searchRange(self, A, target):
        # write your code here
        result = [-1,-1]
        if A is None or len(A) == 0:
            return result

        start = 0
        end = len(A) - 1
        while(start + 1 < end):
            mid = start + (end - start)/2
            if A[mid] >= target:
                end = mid
            elif A[mid] < target:
                start = mid
        if A[start] == target:
            result[0] = start
        elif A[end] == target:
            result[0] = end
        else:
            return result
        start = 0
        end = len(A) - 1
        while(start + 1 < end):
            mid = start + (end - start)/2
            if A[mid] > target:
                end = mid
            elif A[mid] <= target:
                start = mid
        if A[end] == target:
            result[1] = end
            return result
        elif A[start] == target:
            result[1] = start
            return result
        return result