最大字段和问题 C++(穷举、分治法、动态规划)

问题描述

给定由n个整数（包含负整数）组成的序列a1,a2,…,an，求该序列子段和的最大值。规定当所有整数均为负值时定义其最大子段和为0

穷举法

最简单的方法就是穷举法，用一个变量指示求和的开始位置，一个变量指示结束位置，再一个变量指示当前要加和的位置，每一个开始位置对应n-i个结束位置，遍历一遍就能得到最大值

int maxSubArray(int a[], int n) {int maxsum = 0;for (int i = 0; i < n; i++) {//开始位置for (int j = i; j < n; j++) {//结束位置int nowsum = 0;for (int k = i; k <= j; k++) {nowsum += a[k];if (nowsum > maxsum)maxsum = nowsum;}}}return maxsum;
}

算法有三重循环，时间复杂性为O(n^3)

穷举法优化
当字段的开始下标确定后，要计算[i:j]的字段和可以利用上一次计算的[i:j-1]的字段和，加上a[j]就可以了

int maxSubArray2(int a[], int n) {int maxsum = 0;for (int i = 0; i < n; i++) {//开始位置int nowsum = 0;for (int j = i; j < n; j++) {//结束位置nowsum += a[j];if (nowsum > maxsum)maxsum = nowsum;}}return maxsum;
}

改进后的时间复杂度为O(n^2)

分治法

该问题也可以用分治法解决

分治策略思想如下:
将所给序列a[1:n]分成长度相同的两端a[1:n/2]、a[n/2 +1:n]，分别求出这两段的最大子段和，则整体序列a[1:n]的最大子段和有以下三种情况

a[1:n]的最大子段和与a[1:n/2]的最大子段和相同
a[1:n]的最大子段和与a[n/2 +1:n]的最大子段和相同
a[1:n]的最大子段和是a[1:n/2]最后一段加a[n/2 +1:n]最开始一段的和

前两种情况可以递归求得。对于第三种情况，可以发现，a[n/2]和a[n/2 +1]都在最大子段里，我们可以从a[n/2]向左、从a[n/2 +1]向右分别计算两个最大字段和s1和s2，s1+s2就是最大子段和

递归方程

$\text{MaxSum}(low, high) = \begin{cases} \displaystyle \max\left(0,\, \text{arr}[low]\right) & \text{if } low = high \\ \displaystyle \max \begin{cases} \text{MaxSum}(low, mid) \\ \text{MaxSum}(mid+1, high) \\ \text{CrossSum}(low, mid, high) \end{cases} & \text{otherwise} \end{cases}$

代码

int maxSubArray3(int a[], int left, int right) {if (left == right)return a[left];int mid = (left + right) / 2;int maxleft = maxSubArray3(a, left, mid);int maxright = maxSubArray3(a, mid + 1, right);int maxleftsum = 0, maxrightsum = 0;int nowsum = 0;for (int i = mid; i >= left; i--) {nowsum += a[i];if (nowsum > maxleftsum)maxleftsum = nowsum;}nowsum = 0;for (int i = mid + 1; i <= right; i++) {nowsum += a[i];if (nowsum > maxrightsum)maxrightsum = nowsum;}return max(maxleft, max(maxright, maxleftsum + maxrightsum));
}

$T(n)=\big \{^{O(1) \quad n\le c}_{2T(\frac{n}{2})+O(n) \quad n>c}$
根据master定理，我们得到
$T (n) = O (n l o g)$

比起穷举法的O(n^2)更优了

动态规划

设数组为 a[1..n]，定义状态 b[i] 表示以 a[i] 结尾的子段的最大和，则最大字段和为 $max b_j$

有递归方程：
$\begin{cases} 0 & i = 0 \quad (\text{边界条件}) \\ \max(b[i-1] + a[i],\, a[i]) & i \ge 1 \end{cases}$
全局最大子段和为所有 b[i] 中的最大值，并与0比较：
$\text{MaxSum} = \max\left(0,\, \max_{1 \le i \le n} b[i]\right)$

计算最优值

使用变量b记录此前最大字段和，如果为负数则当前最大和为a[i]，如果为正数则最大和为b+a[i]
代码

int maxSubArray4(int a[], int n) {int b = 0;int maxsum = 0;for (int i = 0; i < n; i++) {if (b > 0)b += a[i];else b = a[i];if (b > maxsum)maxsum = b;}return maxsum;
}

构造最优解

使用两个变量start和end记录最大字段的起始和结束位置

int maxSubArray4(int a[], int n, int* start, int* end) {int b = 0;int max_sum = 0;int current_start = 0;  // 当前子段起始位置*start = *end = -1;     // 默认无效索引for (int i = 0; i < n; i++) {if (b > 0) {b += a[i];}else {b = a[i];current_start = i;  // 重置起点}// 更新全局最大值及索引if (b > max_sum) {max_sum = b;*start = current_start;*end = i;}}// 处理全负数情况：返回0且不记录索引if (max_sum <= 0) {*start = *end = -1;return 0;}return max_sum;
}

时间负责度：O(n)

实例

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
using namespace std;
//穷举法
int maxSubArray(int a[], int n) {int maxsum = 0;for (int i = 0; i < n; i++) {//开始位置for (int j = i; j < n; j++) {//结束位置int nowsum = 0;for (int k = i; k <= j; k++) {nowsum += a[k];if (nowsum > maxsum)maxsum = nowsum;}}}return maxsum;
}
//穷举法优化
int maxSubArray2(int a[], int n) {int maxsum = 0;for (int i = 0; i < n; i++) {//开始位置int nowsum = 0;for (int j = i; j < n; j++) {//结束位置nowsum += a[j];if (nowsum > maxsum)maxsum = nowsum;}}return maxsum;
}
//分治法
int maxSubArray3(int a[], int left, int right) {if (left == right)return a[left];int mid = (left + right) / 2;int maxleft = maxSubArray3(a, left, mid);int maxright = maxSubArray3(a, mid + 1, right);int maxleftsum = 0, maxrightsum = 0;int nowsum = 0;for (int i = mid; i >= left; i--) {nowsum += a[i];if (nowsum > maxleftsum)maxleftsum = nowsum;}nowsum = 0;for (int i = mid + 1; i <= right; i++) {nowsum += a[i];if (nowsum > maxrightsum) {maxrightsum = nowsum;}}return max(maxleft, max(maxright, maxleftsum + maxrightsum));
}
//动态规划
int maxSubArray4(int a[], int n, int* start, int* end) {int b = 0;int max_sum = 0;int current_start = 0;  // 当前子段起始位置*start = *end = -1;     // 默认无效索引for (int i = 0; i < n; i++) {if (b > 0) {b += a[i];}else {b = a[i];current_start = i;  // 重置起点}// 更新全局最大值及索引if (b > max_sum) {max_sum = b;*start = current_start;*end = i;}}// 处理全负数情况：返回0且不记录索引if (max_sum <= 0) {*start = *end = -1;return 0;}return max_sum;
}
int main() {int a[] = { 1, -2, 3, 10, -4, 7, 2, -5 };cout << maxSubArray(a, 8) << endl;cout << maxSubArray2(a, 8) << endl;cout << maxSubArray3(a, 0, 7) << endl;int start, end;cout << maxSubArray4(a, 8, &start, &end) << endl;cout <<"start:" << start << " " <<"end:"<< end << endl;return 0;
}