最大子數組問題（分治法）--【演算法導論】

動態規劃解決這個問題

，具體內容我不多說，轉的內容是：

13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7

1.完全位於子數組A[low..mid]中， 因此low<=i<=j<=mid;

2.完全位於子數組A[mid + 1..high]中，因此mid<=i<=j<=high;

3.跨越了中點，因此low<=i<=mid<j<=high;

中或者跨越中點的所有子數組中和最大者。

代碼：

#include <iostream>const int Infinite = -10000;//int max_left = 0;//int max_right = 0;using namespace std;int FindMaxCrossSubarray(int A[], int low, int mid, int high)  //跨越{    int left_sum = Infinite;    int sum = 0;    for (int i = mid; i >= low; i--)  //左半部的最大子數組    {        sum += A[i];        if (sum >left_sum)        {            left_sum = sum;            //max_left = i;        }    }    int right_sum = Infinite;    sum = 0;    for (int i = mid + 1; i <= high; i++)  //右半部的最大子數組    {        sum += A[i];        if (sum > right_sum)        {            right_sum = sum;            //max_right = i;        }    }    return left_sum + right_sum;}int FindMaxSubarray(int A[], int low, int high){    int left_sum, right_sum, cross_sum;    if (high == low)  //一個元素    {        return A[low];    }    else    {        int mid = (low + high) / 2; //分治        left_sum = FindMaxSubarray(A, low, mid);  //前半部        right_sum = FindMaxSubarray(A, mid + 1, high);  //後半部        cross_sum = FindMaxCrossSubarray(A, low, mid, high);  //跨越前後        if (left_sum >= right_sum && left_sum >= cross_sum)  //最大子數組在左邊            return left_sum;        else if (right_sum >= left_sum && right_sum >= cross_sum)  //右邊            return right_sum;        else  //跨越            return cross_sum;    }}int main(){    int a[] = {13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7};    //int a[] = {23, -4};    int length = sizeof(a) / sizeof(int);    cout<<FindMaxSubarray(a, 0, length - 1)<<endl;    //cout<<"最大子序列的下標:"<<max_left<<"->"<<max_right;    return 0;}

 

歡迎指點，o(∩_∩)o