刷题-滑动窗口类型 求连续数组最大值

1. Pattern: Sliding window，滑动窗口类型
滑动窗口类型的题目经常是用来执行数组或是链表上某个区间（窗口）上的操作。比如找最长的全为1的子数组长度。滑动窗口一般从第一个元素开始，
一直往右边一个一个元素挪动。当然了，根据题目要求，我们可能有固定窗口大小的情况，也有窗口的大小变化的情况。

该图中，我们的窗子不断往右一格一个移动
下面是一些我们用来判断我们可能需要上滑动窗口策略的方法：
这个问题的输入是一些线性结构：比如链表呀，数组啊，字符串啊之类的
让你去求最长/最短子字符串或是某些特定的长度要求
经典题目：
Maximum Sum Subarray of Size K (easy)

/*
查找最大子数组和
Find maximum(or minimum) sum of a subarray of size k
Last Updated : 09 - 05 - 2020
Given an array of integers and a number k, find maximum sum of a subarray of size k.

Examples :

Input: arr[] = { 100, 200, 300, 400 }
    k = 2
    Output : 700

    Input : arr[] = { 1, 4, 2, 10, 23, 3, 1, 0, 20 }
    k = 4
    Output : 39
    We get maximum sum by adding subarray{ 4, 2, 10, 23 }
    of size 4.

    Input : arr[] = { 2, 3 }
    k = 3
    Output : Invalid
    There is no subarray of size 3 as size of whole
    array is 2.
*/
//O(n) solution for finding maximum sum of a subarray of size k
#include <iostream>
using namespace std;

//Return s maximum sum in a subarray of size k.
int maxSum(int arr[], int n, int k)
{
    // k must be greater 
    if (n < k)
    {
        cout << "Invalid";
        return -1;
    }

    //Compute sum of first window of size k
    int res = 0;
    for (int i = 0; i < k; i++)
        res += arr[i];
    //Compute sums of remaining windows by removing first element of previous 
    //window and adding last element of current window.
    int curr_sum = res;
    for (int i = k; i < n; i++) {
        curr_sum += arr[i] - arr[i - k];
        res = max(res, curr_sum);
    }
    return res;
}

//Drier code
int main() {
    int arr[] = { 1,4,2,10,2,3,1,0,20 };
    int k = 4;
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maxSum(arr, n, k);
    return 0;
}


/*
求出所有的连续子数组
Subarray/Substring vs Subsequence and Programs to Generate them
Last Updated: 18-09-2018
Subarray/Substring
A subbarray is a contiguous part of array. An array that is 
inside another array. For example, consider the array [1, 2, 3, 4], 
There are 10 non-empty sub-arrays. 
The subbarays are (1), (2), (3), (4), (1,2), (2,3), (3,4), (1,2,3), (2,3,4) and (1,2,3,4). 
In general, for an array/string of size n, there are n*(n+1)/2 non-empty subarrays/subsrings.

All Non-empty Subarrays
1
1 2
1 2 3
1 2 3 4
2
2 3
2 3 4
3
3 4
4
*/

#include <stdio.h>
using namespace std;
//print all subarrays in arr[0...n-1]
void subArray(int arr[], int n)
{
    //pick start point
    for (int i = 0; i < n; i++)
    {
        //pick ending point
        for(int j = i; j < n; j++)
        {
            //print subarray between current starting
            for (int k = j; k <= j; k++) {
                cout << arr[k] << ",";
            }
            cout << endl;
        }
    }
}

//Driver program
int main()
{
    int arr[] = { 1,2,3,4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "All Non-empty Subarray\n";
    subArray(arr,n);
    return 0;
}

Smallest Subarray with a given sum (easy)
Longest Substring with K Distinct Characters (medium)
Fruits into Baskets (medium)
No-repeat Substring (hard)
Longest Substring with Same Letters after Replacement (hard)
Longest Subarray with Ones after Replacement (hard)
一，Maximum Sum Subarray of Size K (easy)
Given an array arr of size N and an integer K, the task is to find the maximum sum subarray of size k among all contiguous sub-array (considering circular subarray also).
Examples:Input: arr = {18, 4, 3, 4, 5, 6, 7, 8, 2, 10}, k = 3
Output:
max circular sum = 32，start index = 9，end index = 1，Explanation:Maximum Sum = 10 + 18 + 4 = 32
解答1：
//C++ program to find maximum circular
//subarray sum of size k
#include <bits/stdc++.h>
using namespace std;
//Function to calculate
//maximum sum
void maxCircularSum(int arr[], int n, int k)
{
    //k must be greater
    if (n < k) {
        cout << "Invalid";
        return;
    }
    int sum = 0, start = 0, end = k - 1;
    //calculate the sum of first k elements;
    for (int i = 0; i < k; i++)
    {
        sum += arr[i];
    }
    int ans = sum;
    for (int i = k; i < n + k; i++)
    {
        //add current element to sum
        //and substract the first element
        //of the previous window.
        sum += arr[i % n] - arr[(i - k) % n];
        if (sum > ans) {
            ans = sum;
            start = (i - k + 1) % n;
            end = i % n;
        }
    }
    cout << "Max circular sum=" << ans << endl;
    cout << "start index=" << start << "\nend index=" << end << endl;
}
//Driver Code
int main()
{
    int arr[] = { 18,4,3,4,5,6,7,8,2,10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
    maxCircularSum(arr,n,k);
    return 0;
}