[leetcode]Search for a Range

二分查找。找到后往左往右移。又说有种log(n)的算法是二分查找最左的，然后二分查找最右的。省却最后那部分O(n)的

public class Solution {

    public int[] searchRange(int[] A, int target) {

        // Start typing your Java solution below

        // DO NOT write main() function

        int pos = searchTarget(A, target, 0, A.length -1);

        if (pos == -1) return new int[]{-1,-1};

        

        int start = 0;

        int end = A.length - 1;

        for (int i = pos - 1; i >= 0; i--) {

            if (A[i] != target) {

        		start = i+1;

        		break;

        	}       		

        }

        for (int i = pos + 1; i < A.length; i++) {

        	if (A[i] != target) {

        		end = i - 1;

        		break;

        	}

        }

        return new int[]{start,end};

    }

    

    public int searchTarget(int[] A, int target, int left, int right) {

    	if (left > right) return -1;

    	int mid = (left + right) / 2;

    	if (A[mid] == target) return mid;

    	if (A[mid] > target) return searchTarget(A, target, left, mid -1);

    	else return searchTarget(A, target, mid+1, right);   	

    }

}

这是参考答案，两次log(n)的查找：http://discuss.leetcode.com/questions/213/search-for-a-range

class Solution {

public:

    vector<int> searchRange(int A[], int n, int target) {

        vector<int> range(2, -1);

        int lower = 0;

        int upper = n;

        int mid;



        // Search for lower bound

        while (lower < upper) {

            mid = (lower + upper) / 2;

            if (A[mid] < target)

                lower = mid + 1;

            else

                upper = mid;

        }



        // If the target is not found, return (-1, -1)

        if (A[lower] != target)

            return range;

        range[0] = lower;



        // Search for upper bound

        upper = n;

        while (lower < upper) {

            mid = (lower + upper) / 2;

            if (A[mid] > target)

                upper = mid;

            else

                lower = mid + 1;

        }

        range[1] = upper - 1;



        return range;

    }

};