zoukankan      html  css  js  c++  java
  • LeetCode(35) Search Insert Position

    题目

    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

    分析

    此题目与上一个题目类似都是源于二分查找的变形。
    若序列中存在目标元素值,则直接返回其下标,若不存在则返回第一个大于它的元素的下标,只需要在二分搜索算法中稍微修改即可。

    AC代码

    class Solution {
    public:
        int searchInsert(vector<int>& nums, int target) {
            if (nums.size() == 0)
                return 0;
            else if (nums.size() == 1)
            {
                if (nums[0] >= target)
                    return 0;
                else
                    return 1;
            }
            else{
                return BinarySearch(nums, target);
            }
        }
    
        int BinarySearch(vector<int> & nums, int target)
        {
            int left = 0, right = nums.size() - 1;
    
            while (left <= right)
            {
                int mid = (left + right) / 2;
                if (nums[mid] == target)
                    return mid;
                else if (nums[mid] < target)
                {
                    if (mid == right || nums[mid + 1] > target)
                        return mid + 1;
                    else
                        left = mid + 1;
                }
                else{
                    if (mid == left || nums[mid - 1] < target)
                        return mid;
                    else
                        right = mid - 1;
                }
            }//while
    
            return -1;
        }
    };
    

    GitHub测试程序源码

  • 相关阅读:
    3.13作业 制作网页布局
    3.11 框架和样式表
    表单
    3.8学习记录
    第一次作业
    数据库增删改查
    数据库三大范式
    数据库中的时间戳
    数据库的主键与外键
    登录页面
  • 原文地址:https://www.cnblogs.com/shine-yr/p/5214907.html
Copyright © 2011-2022 走看看