LC刷题第四天-优快云博客

本文链接：https://blog.youkuaiyun.com/MATLAB2020ab/article/details/124314580

本文介绍了招商银行信用卡的相关内容，并围绕编程技术展开，包括求最大子数组和的动态规划、整数反转的位操作技巧，以及在旋转排序数组中进行搜索的二分查找策略。通过实例展示了Solution类中的三个核心算法实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

招商银行信用卡

53. 最大子数组和

// shiran
#include <bits/stdc++.h>
using namespace std;

#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, n, a) for (int i = n - 1; i >= a; i--)
#define sz(x) (int)size(x)
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define pb push_back
typedef long long ll;
typedef pair<int, int> PII;
const int mod = 1000000007;
const int N = 1010;
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

class Solution
{
public:
    int maxSubArray(vector<int> &nums)
    {
        int a = nums[0], res = nums[0];
        int n = sz(nums);
        rep(i, 1, n)
            a = max(nums[i], a + nums[i]),
            res = max(res, a);
        return res;
    }
};

7. 整数反转

// shiran
#include <bits/stdc++.h>
using namespace std;

#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, n, a) for (int i = n - 1; i >= a; i--)
#define sz(x) (int)size(x)
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define pb push_back
typedef long long ll;
typedef pair<int, int> PII;
const int mod = 1000000007;
const int N = 1010;
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

class Solution
{
public:
    int reverse(int x)
    {
        ll t = x, ans = 0;
        if (t < 0)
        {
            t = -t;
            while (t)
            {
                ans = ans * 10 + t % 10;
                t /= 10;
            }
            ans = -ans;
            if (ans < INT_MIN)
                return 0;
            else
                return ans;
        }
        else
        {
            while (t)
            {
                ans = ans * 10 + t % 10;
                t /= 10;
            }
            if (ans > INT_MAX)
                return 0;
            return ans;
        }
        return 0;
    }
};

33. 搜索旋转排序数组

先二分找到中间的分界点，然后再二分找到 $t a r g e t$

// shiran
#include <bits/stdc++.h>
using namespace std;

#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, n, a) for (int i = n - 1; i >= a; i--)
#define sz(x) (int)size(x)
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define pb push_back
typedef long long ll;
typedef pair<int, int> PII;
const int mod = 1000000007;
const int N = 1010;
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

class Solution
{
public:
    int search(vector<int> &nums, int target)
    {
        int l = 0, r = nums.size() - 1;
        while (l < r)
        {
            int mid = l + r + 1 >> 1;
            if (nums[mid] >= nums[0])
                l = mid;
            else
                r = mid - 1;
        }

        if (target >= nums[0])
            l = 0;
        else
            l = r + 1, r = nums.size() - 1;
        while (l < r)
        {
            int mid = l + r >> 1;
            if (nums[mid] >= target)
                r = mid;
            else
                l = mid + 1;
        }
        if (nums[r] == target)
            return r;
        else
            return -1;
    }
};