给定一个整数数组arr[] ，任务是在所有可能的子数组中找到最大和的子数组。
例子：
Input: arr[] = {-2, 1, -3, 4, -1, 2, 1, -5, 4}
Output: 6
{4, -1, 2, 1} is the required sub-array.
Input: arr[] = {2, 2, -2}
Output: 4
方法：到目前为止，我们只知道 Kadane 算法，它使用动态规划在 O(n) 中解决了这个问题。
我们还讨论了 O(N*logN) 时间复杂度的最大和子阵列的分治法。
下面的方法使用分而治之的方法来解决它，该方法的时间复杂度与O(n) 相同。
分治算法通常涉及将问题划分为子问题并分别解决它们。对于这个问题，我们维护一个结构(在 cpp 中)或类(在Java或Python)，它存储以下值：
- 子数组的总和。
- 子数组的最大前缀和。
- 子数组的最大后缀和。
- 子数组的总体最大和。(这包含子数组的最大和)。
在递归(除部分)过程中，数组从中间分成 2 部分。左节点结构包含数组左部分的所有上述值，右节点结构包含上述所有值。拥有两个节点，现在我们可以通过计算结果节点的所有值来合并两个节点。
结果节点的最大前缀和将是左节点的最大前缀和或左节点的最大前缀和 + 右节点的最大前缀和或两个节点的总和中的最大值(这对于具有所有正值的数组是可能的) .
类似地，结果节点的最大后缀和将是右节点的最大后缀和或右节点的最大后缀和 + 左节点的最大后缀和或两个节点的总和中的最大值(这对于所有为正的数组也是可能的)值)。
结果节点的总和是左节点和右节点总和的总和。
现在，结果节点的最大子数组和将在结果节点的前缀和、结果节点的后缀和、结果节点的总和、左节点的最大和、右节点的最大和、最大后缀和的总和中最大左节点和右节点的最大前缀和。
这里征服部分可以通过组合左右节点结构的结果在 O(1) 时间内完成。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
struct Node {
// To store the maximum sum
// for a sub-array
long long _max;
// To store the maximum prefix
// sum for a sub-array
long long _pre;
// To store the maximum suffix
// sum for a sub-array
long long _suf;
// To store the total sum
// for a sub-array
long long _sum;
};
// Function to create a node
Node getNode(long long x){
Node a;
a._max = x;
a._pre = x;
a._suf = x;
a._sum = x;
return a;
}
// Function to merge the 2 nodes left and right
Node merg(const Node &l, const Node &r){
// Creating node ans
Node ans ;
// Initializing all the variables:
ans._max = ans._pre = ans._suf = ans._sum = 0;
// The max prefix sum of ans Node is maximum of
// a) max prefix sum of left Node
// b) sum of left Node + max prefix sum of right Node
// c) sum of left Node + sum of right Node
ans._pre = max({l._pre, l._sum+r._pre, l._sum+r._sum});
// The max suffix sum of ans Node is maximum of
// a) max suffix sum of right Node
// b) sum of right Node + max suffix sum of left Node
// c) sum of left Node + sum of right Node
ans._suf = max({r._suf, r._sum+l._suf, l._sum+r._sum});
// Total sum of ans Node = total sum of left Node + total sum of right Node
ans._sum = l._sum + r._sum;
// The max sum of ans Node stores the answer which is the maximum value among:
// prefix sum of ans Node
// suffix sum of ans Node
// maximum value of left Node
// maximum value of right Node
// prefix value of right Node + suffix value of left Node
ans._max = max({ans._pre, ans._suf, ans._sum,l._max, r._max, l._suf+r._pre});
// Return the ans Node
return ans;
}
// Function for calculating the
// max_sum_subArray using divide and conquer
Node getMaxSumSubArray(int l, int r, vector &ar){
if (l == r) return getNode(ar[l]);
int mid = (l + r) >> 1;
// Call method to return left Node:
Node left = getMaxSumSubArray(l, mid, ar);
// Call method to return right Node:
Node right = getMaxSumSubArray(mid+1, r, ar);
// Return the merged Node:
return merg(left, right);
}
// Driver code
int main(){
vector ar = {-2, -5, 6, -2, -3, 1, 5, -6};
int n = ar.size();
Node ans = getMaxSumSubArray(0, n-1, ar);
cout << "Answer is " << ans._max << "\n";
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static class Node
{
// To store the maximum sum
// for a sub-array
int _max;
// To store the maximum prefix
// sum for a sub-array
int _pre;
// To store the maximum suffix
// sum for a sub-array
int _suf;
// To store the total sum
// for a sub-array
int _sum;
};
// Function to create a node
static Node getNode(int x)
{
Node a = new Node();
a._max = x;
a._pre = x;
a._suf = x;
a._sum = x;
return a;
}
// Function to merge the 2 nodes left and right
static Node merg(Node l, Node r)
{
// Creating node ans
Node ans = new Node();
// Initializing all the variables:
ans._max = ans._pre = ans._suf = ans._sum = 0;
// The max prefix sum of ans Node is maximum of
// a) max prefix sum of left Node
// b) sum of left Node + max prefix sum of right Node
// c) sum of left Node + sum of right Node
ans._pre = Arrays.stream(new int[]{l._pre, l._sum+r._pre,
l._sum+r._sum}).max().getAsInt();
// The max suffix sum of ans Node is maximum of
// a) max suffix sum of right Node
// b) sum of right Node + max suffix sum of left Node
// c) sum of left Node + sum of right Node
ans._suf = Arrays.stream(new int[]{r._suf, r._sum+l._suf,
l._sum+r._sum}).max().getAsInt();
// Total sum of ans Node = total sum of
// left Node + total sum of right Node
ans._sum = l._sum + r._sum;
// The max sum of ans Node stores
// the answer which is the maximum value among:
// prefix sum of ans Node
// suffix sum of ans Node
// maximum value of left Node
// maximum value of right Node
// prefix value of right Node + suffix value of left Node
ans._max = Arrays.stream(new int[]{ans._pre,
ans._suf,
ans._sum,
l._max, r._max,
l._suf+r._pre}).max().getAsInt();
// Return the ans Node
return ans;
}
// Function for calculating the
// max_sum_subArray using divide and conquer
static Node getMaxSumSubArray(int l, int r, int []ar)
{
if (l == r) return getNode(ar[l]);
int mid = (l + r) >> 1;
// Call method to return left Node:
Node left = getMaxSumSubArray(l, mid, ar);
// Call method to return right Node:
Node right = getMaxSumSubArray(mid + 1, r, ar);
// Return the merged Node:
return merg(left, right);
}
// Driver code
public static void main(String[] args)
{
int []ar = {-2, -5, 6, -2, -3, 1, 5, -6};
int n = ar.length;
Node ans = getMaxSumSubArray(0, n - 1, ar);
System.out.print("Answer is " + ans._max + "\n");
}
}
// This code is contributed by shikhasingrajputPython3
# Python3 implementation of the approach
class Node:
def __init__(self, x):
# To store the maximum sum for a sub-array
self._max = x
# To store the maximum prefix sum for a sub-array
self._pre = x
# To store the maximum suffix sum for a sub-array
self._suf = x
# To store the total sum for a sub-array
self._sum = x
# Function to merge the 2 nodes left and right
def merg(l, r):
# Creating node ans
ans = Node(0)
# The max prefix sum of ans Node is maximum of
# a) max prefix sum of left Node
# b) sum of left Node + max prefix sum of right Node
# c) sum of left Node + sum of right Node
ans._pre = max(l._pre, l._sum+r._pre, l._sum+r._sum)
# The max suffix sum of ans Node is maximum of
# a) max suffix sum of right Node
# b) sum of right Node + max suffix sum of left Node
# c) sum of left Node + sum of right Node
ans._suf = max(r._suf, r._sum+l._suf, l._sum+r._sum)
# Total sum of ans Node = total sum of
# left Node + total sum of right Node
ans._sum = l._sum + r._sum
# The max sum of ans Node stores the answer
# which is the maximum value among:
# prefix sum of ans Node
# suffix sum of ans Node
# maximum value of left Node
# maximum value of right Node
# prefix value of left Node + suffix value of right Node
ans._max = max(ans._pre, ans._suf, ans._sum,
l._max, r._max, l._suf+r._pre)
# Return the ans Node
return ans
# Function for calculating the
# max_sum_subArray using divide and conquer
def getMaxSumSubArray(l, r, ar):
if l == r: return Node(ar[l])
mid = (l + r) // 2
# Call method to return left Node:
left = getMaxSumSubArray(l, mid, ar)
# Call method to return right Node:
right = getMaxSumSubArray(mid+1, r, ar)
# Return the merged Node:
return merg(left, right)
# Driver code
if __name__ == "__main__":
ar = [-2, -5, 6, -2, -3, 1, 5, -6]
n = len(ar)
ans = getMaxSumSubArray(0, n-1, ar)
print("Answer is", ans._max)
# This code is contributed by Rituraj JainC#
// C# implementation of the approach
using System;
using System.Linq;
public class GFG
{
class Node
{
// To store the maximum sum
// for a sub-array
public int _max;
// To store the maximum prefix
// sum for a sub-array
public int _pre;
// To store the maximum suffix
// sum for a sub-array
public int _suf;
// To store the total sum
// for a sub-array
public int _sum;
};
// Function to create a node
static Node getNode(int x)
{
Node a = new Node();
a._max = x;
a._pre = x;
a._suf = x;
a._sum = x;
return a;
}
// Function to merge the 2 nodes left and right
static Node merg(Node l, Node r)
{
// Creating node ans
Node ans = new Node();
// Initializing all the variables:
ans._max = ans._pre = ans._suf = ans._sum = 0;
// The max prefix sum of ans Node is maximum of
// a) max prefix sum of left Node
// b) sum of left Node + max prefix sum of right Node
// c) sum of left Node + sum of right Node
ans._pre = (new int[]{l._pre, l._sum+r._pre,
l._sum+r._sum}).Max();
// The max suffix sum of ans Node is maximum of
// a) max suffix sum of right Node
// b) sum of right Node + max suffix sum of left Node
// c) sum of left Node + sum of right Node
ans._suf = (new int[]{r._suf, r._sum+l._suf,
l._sum+r._sum}).Max();
// Total sum of ans Node = total sum of
// left Node + total sum of right Node
ans._sum = l._sum + r._sum;
// The max sum of ans Node stores
// the answer which is the maximum value among:
// prefix sum of ans Node
// suffix sum of ans Node
// maximum value of left Node
// maximum value of right Node
// prefix value of right Node + suffix value of left Node
ans._max = (new int[]{ans._pre,
ans._suf,
ans._sum,
l._max, r._max,
l._suf+r._pre}).Max();
// Return the ans Node
return ans;
}
// Function for calculating the
// max_sum_subArray using divide and conquer
static Node getMaxSumSubArray(int l, int r, int []ar)
{
if (l == r) return getNode(ar[l]);
int mid = (l + r) >> 1;
// Call method to return left Node:
Node left = getMaxSumSubArray(l, mid, ar);
// Call method to return right Node:
Node right = getMaxSumSubArray(mid + 1, r, ar);
// Return the merged Node:
return merg(left, right);
}
// Driver code
public static void Main(String[] args)
{
int []ar = {-2, -5, 6, -2, -3, 1, 5, -6};
int n = ar.Length;
Node ans = getMaxSumSubArray(0, n - 1, ar);
Console.Write("Answer is " + ans._max + "\n");
}
}
// This code is contributed by shikhasingrajput输出：
Answer is 7时间复杂度： getMaxSumSubArray() 递归函数生成以下递归关系。
T(n) = 2 * T(n / 2) + O(1)注意征服部分只需要 O(1) 时间。因此，在使用 Master 定理求解此递归时，我们得到 O(n) 的时间复杂度。
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