leetCode-Search Insert Position

最新推荐文章于 2024-03-25 23:05:15 发布

weixin_30381793

最新推荐文章于 2024-03-25 23:05:15 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/kevincong/p/7803407.html

本文介绍了两种二分查找算法的实现方法：递归版本和非递归版本。这两种方法都用于在一个已排序的数组中找到目标值的位置，如果目标值不存在，则返回其应有的插入位置。递归版本使用了辅助函数进行递归调用，而非递归版本通过循环来迭代搜索范围。

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;
        }
    }
}