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文章目錄

• 前言
• 一、動態規劃法
• 二、分治法
• 三、程序總代碼
• 總結

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前言

本次將對最大字段和使用C語言實現的兩種方法實現，動態規劃法和分治法，兩種方法所反映和表現是不同的想法和做法，但是動態規劃法的時間復雜度更低，那并不是說分治法就不行，相反只是在解答最大字段和的時候沒有動態規劃法來的高效罷了。代碼都比較的好理解，認真看都能看得懂，這里就不解釋了！

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一、動態規劃法

int maxsum2(int a[],int left,int right) {int first = left + 1;int total = right - left + 1;int b[100];int sum = 0;b[left] = a[left];int i, j;for (i = first;i <= right;i++){b[i]=(b[i-1] + a[i]) > 0 ? (b[i-1] + a[i]) : a[i];}for (j = left;j <= right;j++){sum = sum > b[j] ? sum : b [j] ;}return sum; }

二、分治法

int maxsum1(int a[], int left, int right) {int leftsum, rightsum, midsum;int mid;int i, j;int temporary1=0, temporary2=0;int sum1=0, sum2=0,sum;if (left == right){sum = a[left];}else {mid = (left + right) / 2;leftsum = maxsum1(a, left, mid);rightsum = maxsum1(a,mid+1,right);for (i = mid;i >= left;i--){sum1 += a[i];temporary1 = sum1 > temporary1 ? sum1:temporary1;}for (j = mid + 1;j <= right;j++){sum2 += a[j];temporary2 = sum2 > temporary2 ? sum2 : temporary2;}midsum = temporary1 + temporary2;if (leftsum > rightsum ){sum = leftsum;}else {sum = rightsum;}if (sum < midsum){sum = midsum;}}return sum; }

三、程序總代碼

#include<stdio.h> int maxsum1(int a[], int left, int right); //分治法求最大字段和 int maxsum2(int a[], int left, int right); //動態規劃法求最大字段和 int main() {int a[10];int max;for (int i = 0;i < 10;i++){scanf("%d", &a[i]);}for (int j = 0;j < 10;j++){printf("%d ", a[j]);}//max=maxsum1(a, 0, 9);max = maxsum2(a,0,9);printf("\n%d", max);return 0; } int maxsum1(int a[], int left, int right) {int leftsum, rightsum, midsum;int mid;int i, j;int temporary1=0, temporary2=0;int sum1=0, sum2=0,sum;if (left == right){sum = a[left];}else {mid = (left + right) / 2;leftsum = maxsum1(a, left, mid);rightsum = maxsum1(a,mid+1,right);for (i = mid;i >= left;i--){sum1 += a[i];temporary1 = sum1 > temporary1 ? sum1:temporary1;}for (j = mid + 1;j <= right;j++){sum2 += a[j];temporary2 = sum2 > temporary2 ? sum2 : temporary2;}midsum = temporary1 + temporary2;if (leftsum > rightsum ){sum = leftsum;}else {sum = rightsum;}if (sum < midsum){sum = midsum;}}return sum; }int maxsum2(int a[],int left,int right) {int first = left + 1;int total = right - left + 1;int b[100];int sum = 0;b[left] = a[left];int i, j;for (i = first;i <= right;i++){b[i]=(b[i-1] + a[i]) > 0 ? (b[i-1] + a[i]) : a[i];}for (j = left;j <= right;j++){sum = sum > b[j] ? sum : b [j] ;}return sum; }

總結

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總結

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