leetcode 34. Search for a Range

Given an array of integers sorted in ascending order, 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].

做两次二分查找，寻找左右的边界

不过自己最后有点绕进去了，因为两个相邻的数相加数/2总是偏小的那个数。。。 = = 

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] ans = new int[]{-1, -1};
        if(nums==null || nums.length==0) return ans;
        int left = 0 , right = nums.length-1;
        while(left<right){
            int mid = (left+right)/2;
            if(nums[mid]<target) left = mid + 1;
            else right = mid;
        }
        if(nums[left]!=target)  return ans;
        ans[0] = left;
        left = 0;
        right = nums.length-1;
        while(left<right){
            int mid = (left+right)/2;
            if(nums[mid]<=target) left = mid + 1;
            else right = mid - 1;
        }
        if(nums[left]==target) ans[1] = left;
        else ans[1] = left-1;
        return ans;
    }
}

可以在第二个mid求取时+1使其向右偏，可以省下面的判断

另一种求该数位置和该数+1位置的二分法思路

public class Solution {
	public int[] searchRange(int[] A, int target) {
		int start = Solution.firstGreaterEqual(A, target);
		if (start == A.length || A[start] != target) {
			return new int[]{-1, -1};
		}
		return new int[]{start, Solution.firstGreaterEqual(A, target + 1) - 1};
	}

	//find the first number that is greater than or equal to target.
	//could return A.length if target is greater than A[A.length-1].
	//actually this is the same as lower_bound in C++ STL.
	private static int firstGreaterEqual(int[] A, int target) {
		int low = 0, high = A.length;
		while (low < high) {
			int mid = low + ((high - low) >> 1);
			//low <= mid < high
			if (A[mid] < target) {
				low = mid + 1;
			} else {
				//should not be mid-1 when A[mid]==target.
				//could be mid even if A[mid]>target because mid<high.
				high = mid;
			}
		}
		return low;
	}
}