Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0

https://oj.leetcode.com/problems/search-insert-position/

思路：binary search 变体，求floor。

public class Solution {	public int searchInsert(int[] A, int target) {		return floor(A, target);	}	private static int floor(int[] a, int target) {		int low = 0;		int high = a.length;		int mid;		while (low < high) {			mid = (low + high) / 2;			if (target > a[mid])				low = mid + 1;			else				high = mid;		}		return low;	}	public static void main(String[] args) {		System.out.println(new Solution().searchInsert(new int[]{1,3,5,6},5));		System.out.println(new Solution().searchInsert(new int[]{1,3,5,6},2));		System.out.println(new Solution().searchInsert(new int[]{1,3,5,6},7));		System.out.println(new Solution().searchInsert(new int[]{1,3,5,6},0));	}}