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.

Example 1:

Input: [1,3,5,6], 5
Output: 2
Example 2:

Input: [1,3,5,6], 2
Output: 1
Example 3:

Input: [1,3,5,6], 7
Output: 4
Example 1:

Input: [1,3,5,6], 0
Output: 0

我的版本(递归二分):

class Solution {    public int searchInsert(int[] nums, int target) {        if(target > nums[nums.length - 1]){            return nums.length;        }else if(target < nums[0]){            return 0;        }        return binarySearch(nums,0,nums.length -1,target);    }    int binarySearch(int[]nums,int first,int last,int target){        int medium = (first + last) / 2;        if(first == medium && nums[medium] != target){            return first + 1;        }        if(target == nums[medium]){            return medium;        }        if(target > nums[medium]){            first = medium;        }        if(target < nums[medium]){            last = medium;        }        return binarySearch(nums,first,last,target);    }}

改进版本(非递归二分):

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