[leetcode] Search Insert Position

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));

	}

}

第二遍记录：

　　注意用坐标inclusive的时候，right初始化是n-1;

public class Solution {
    public int searchInsert(int[] A, int target) {
        int left=0;
        int right=A.length-1;
        while(left<=right){
            int mid=left+(right-left)/2;
            if(target>A[mid]){
                left=mid+1;
            }else{
                right=mid-1;
            }
            
        }
        return left;
    }
}