最大子串行和的演算法分析

// maxsubsequencesum.cpp : 定義控制台應用程式的入口點。

// 求陣列的最大連續子串行和
// 2017/8/3
#include "stdafx.h"
int maxsubsequencesum1(int a, int n);
int maxsubsequencesum2(int a, int n);
int maxsubsequencesum3(int a, int n);
static
int maxsubsum(int a, int, int);
int maxsubsequencesum4(int a, int n);
int main()
;    printf("alg1:%d\n", maxsubsequencesum1(a, 6));
printf("alg2:%d\n", maxsubsequencesum2(a, 6));
printf("alg3:%d\n", maxsubsequencesum3(a, 6));
printf("alg4:%d\n", maxsubsequencesum4(a, 6));
return0;}
//演算法1：時間複雜度o(n3)
int maxsubsequencesum1(int a,int n) 
if (thissum>maxsum)}}
return maxsum;
}//演算法2：時間複雜度o(n2)
int maxsubsequencesum2(int a, int n) }}
return maxsum;
}//演算法3：時間複雜度為o(nlogn)
int maxsubsequencesum3(int a, int n) 
static
int maxsubsum(int a, int left, int right) 
else 
}center = (left + right) / 2;
maxleftsum = maxsubsum(a, left, center);
maxrightsum = maxsubsum(a, center + 1, right);
maxleftbordersum = 0;
leftbordersum = 0;
for ( i = center; i >=left ; i--)
}maxrightbordersum = 0;
rightbordersum = 0;
for ( i = center+1; i < right; i++)
}return (maxleftsum > maxrightsum ? maxleftsum : maxrightsum) > (maxleftbordersum + maxrightbordersum) ?
(maxleftsum > maxrightsum ? maxleftsum : maxrightsum) : (maxleftbordersum + maxrightbordersum);
}//演算法4：時間複雜度為o(n)
//聯機演算法，在任意時刻都能對他已經讀入的資料給出子串行問題的正確答案
int maxsubsequencesum4(int a, int n) 
else
if (thissum < 0) 
}return maxsum;
}

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