Search for a Range

Given a sorted array of n integers, find the starting and ending position of a given target value.

If the target is not found in the array, return [-1, -1].

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

http://www.lintcode.com/en/problem/search-for-a-range/

Discussion

就是Binary Search问题的扩展，如果真的理解了binary search，这个题一点都不难，无非是用二分找出来target第一次出现的地方（就是binary search），和target最后一次出现的地方（跟上一种情况相反，A[mid]==target时向右查找）。

Solution

class Solution {
    /** 
     *@param A : an integer sorted array
     *@param target :  an integer to be inserted
     *return : a list of length 2, [index1, index2]
     */
public:
    vector<int> searchRange(vector<int> &A, int target) {
        // write your code here
        int n = A.size();
        vector<int> result (2, -1);
        if(n == 0) return result;
        int start = 0, end = n-1;
        int mid;
        while(start+1 < end) {
            mid = start + (end-start)/2;
            if(A[mid] == target) {//go to left part to find the first place.
                end = mid;
            } else if(A[mid] > target) {
                end = mid;
            } else {
                start = mid;
            }
        }
        if(A[start] == target) {//check start first (most left)
            result[0] =start;
        } else if (A[end] == target) {
            result[0] =end;
        } else {
            return result;
        }

        start = 0;
        end = n-1;
        while(start+1 < end) {
            mid = start + (end-start)/2;
            if(A[mid] == target) {//go to right part to find the last place
                start = mid;
            } else if(A[mid] > target) {
                end = mid;
            } else {
                start = mid;
            }
        }
        if(A[end] == target) {//check right first, (most right)
            result[1] = end;
        } else if (A[start] == target) {
            result[1] = start;
        } else {
            return result;
        }
        return result;
    }
};