[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;
    }
};