35. Search Insert Position(easy)-优快云博客

本文详细解析了在已排序数组中查找目标值的插入位置的算法。通过二分查找法，该算法能在O(log n)时间内找到目标值应插入的位置，即使数组中不存在该值。文章提供了C++和Java两种语言的实现代码。

Easy

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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 4:

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

Output: 0

C++：

/*
 @Date    : 2019-01-15 20:52:16
 @Author  : 酸饺子 (changzheng300@foxmail.com)
 @Link    : https://github.com/SourDumplings
 @Version : $Id$
*/

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

class Solution
{
public:
    int searchInsert(vector<int>& nums, int target)
    {
        int b = 0, e = nums.size();
        int mi;
        while (b < e)
        {
            mi = (b + e) >> 1;
            target < nums[mi] ? e = mi : b = mi + 1;
        }
        if (b == 0 || nums[b - 1] < target)
        {
            return b;
        }
        else
            return b - 1;
    }
};

Java：

/**
 * @Date    : 2019-01-15 21:24:53
 * @Author  : 酸饺子 (changzheng300@foxmail.com)
 * @Link    : https://github.com/SourDumplings
 * @Version : $Id$
 *
 * https://leetcode.com/problems/search-insert-position/
*/

class Solution
{
    public int searchInsert(int[] nums, int target)
    {
        int b = 0, e = nums.length;
        while (b < e)
        {
            int mi = (b + e) >> 1;
            if (target < nums[mi])
                e = mi;
            else
                b = mi + 1;
        }
        if (b == 0 || nums[b - 1] < target)
        {
            return b;
        }
        else
            return b - 1;
    }
}