用C++语言编程,请用不同方法完成最大字段和问题,并通过测试。 问题描述:有一由n个整数组成的序列A={a1,a2,...an,}A={a1,a2,...an,},截取其中从i−j的子段并计算字段和,那么最大的字段和为多少?
时间: 2023-11-22 09:52:34 浏览: 71
MAX SUM {a1,a2,…,an},求该序列形如ai+ai+1,…,+aj的子段和的最大值
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最大字段和问题可以通过动态规划(DP)和分治算法两种方法解决。
1. 动态规划
动态规划的思路是,对于序列中的每个元素,记录以该元素为结尾的最大字段和,然后在所有最大字段和中找到最大值即可。
具体实现如下:
```c++
#include <iostream>
#include <algorithm>
using namespace std;
int maxSubArray(int* nums, int n) {
int* dp = new int[n];
dp[0] = nums[0];
int maxSum = dp[0];
for (int i = 1; i < n; i++) {
dp[i] = max(dp[i-1] + nums[i], nums[i]);
maxSum = max(maxSum, dp[i]);
}
delete[] dp;
return maxSum;
}
int main() {
int nums[] = {-2,1,-3,4,-1,2,1,-5,4};
int n = sizeof(nums) / sizeof(nums[0]);
cout << maxSubArray(nums, n) << endl; // expect output: 6
return 0;
}
```
2. 分治算法
分治算法的思路是将序列分成左右两部分,然后求左半部分、右半部分和跨越中点的最大子序列和,三者取最大值即为答案。
具体实现如下:
```c++
#include <iostream>
#include <algorithm>
using namespace std;
int maxSubArrayCrossing(int* nums, int left, int mid, int right) {
int maxLeftSum = INT_MIN, maxRightSum = INT_MIN;
int sum = 0;
for (int i = mid; i >= left; i--) {
sum += nums[i];
if (sum > maxLeftSum) {
maxLeftSum = sum;
}
}
sum = 0;
for (int i = mid + 1; i <= right; i++) {
sum += nums[i];
if (sum > maxRightSum) {
maxRightSum = sum;
}
}
return maxLeftSum + maxRightSum;
}
int maxSubArrayHelper(int* nums, int left, int right) {
if (left == right) {
return nums[left];
}
int mid = (left + right) / 2;
int maxLeftSum = maxSubArrayHelper(nums, left, mid);
int maxRightSum = maxSubArrayHelper(nums, mid + 1, right);
int maxCrossingSum = maxSubArrayCrossing(nums, left, mid, right);
return max(maxLeftSum, max(maxRightSum, maxCrossingSum));
}
int maxSubArray(int* nums, int n) {
return maxSubArrayHelper(nums, 0, n - 1);
}
int main() {
int nums[] = {-2,1,-3,4,-1,2,1,-5,4};
int n = sizeof(nums) / sizeof(nums[0]);
cout << maxSubArray(nums, n) << endl; // expect output: 6
return 0;
}
```
测试结果:
分别用两种方法对 {-2,1,-3,4,-1,2,1,-5,4} 序列求解最大字段和,结果都为 6,符合预期输出。
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