给定包含N个数的数列,求最大的子列和(子列必须是连续的嗷)
下面给出4种算法:
#include <iostream> using namespace std; int maxSubseqSum1(int A[], int N); int maxSubseqSum2(int A[], int N); int maxSubseqSum3(int A[], int N); int main() { int A[] = { 1, 3, -5, 3, 12, -5, -9, 23, 1, -5 }; int m1,m2,m3; m1=maxSubseqSum1(A, 10); cout << m1 << endl; m2 = maxSubseqSum2(A, 10); cout << m2<<endl; m3 = maxSubseqSum3(A, 10); cout << m3 << endl; return 0; } int maxSubseqSum1(int A[], int N) { int maxSum = 0, thisSum = 0; int i, j, k; for (i = 0; i < N; i++)//这个循环用来将i遍历整个数组 { for (j = i; j < N; j++)//这个循环用来获得由i开始的连续数列最大和 { thisSum = 0; for (k = i; k <= j; k++) thisSum += A[k];//thisSum用来保存从A[i]开始到A[j]的和 if (thisSum>maxSum)//比较thissum与maxsum的值大小,从而获得连续的最大值 maxSum = thisSum; } } return maxSum; } int maxSubseqSum2(int A[], int N) { int maxSum = 0, thisSum = 0; int i, j; for (i = 0; i < N; i++)//这个循环用来将i遍历整个数组 { thisSum = 0; for (j = i; j < N; j++)//这个循环用来获得由i开始的连续数列最大和 { thisSum += A[j]; if (thisSum>maxSum) maxSum = thisSum; } } return maxSum; } int maxSubseqSum3(int A[], int N)//在线处理 { int maxSum = 0, thisSum = 0; int i; for (i = 0; i < N; i++)//这个循环用来将i遍历整个数组 { thisSum += A[i]; if (thisSum>maxSum)//发现更大和,更新整个结果 maxSum = thisSum; else if (thisSum < 0)//如果thissum=0,则其不可能使后面的值变大,因此将前面的和抛弃,从头来过 thisSum = 0; } return maxSum; }
其时间复杂度分别是T(N)=O(N3),T(N)=O(N2),T(N)=O(N)。下面还有一个分而治之的算法,其时间复杂度为T(N)=O(NlogN)
int Max3( int A, int B, int C ) { /* 返回3个整数中的最大值 */ return A > B ? A > C ? A : C : B > C ? B : C; } int DivideAndConquer( int List[], int left, int right ) { /* 分治法求List[left]到List[right]的最大子列和 */ int MaxLeftSum, MaxRightSum; /* 存放左右子问题的解 */ int MaxLeftBorderSum, MaxRightBorderSum; /*存放跨分界线的结果*/ int LeftBorderSum, RightBorderSum; int center, i; if( left == right ) { /* 递归的终止条件,子列只有1个数字 */ if( List[left] > 0 ) return List[left]; else return 0; } /* 下面是"分"的过程 */ center = ( left + right ) / 2; /* 找到中分点 */ /* 递归求得两边子列的最大和 */ MaxLeftSum = DivideAndConquer( List, left, center ); MaxRightSum = DivideAndConquer( List, center+1, right ); /* 下面求跨分界线的最大子列和 */ MaxLeftBorderSum = 0; LeftBorderSum = 0; for( i=center; i>=left; i-- ) { /* 从中线向左扫描 */ LeftBorderSum += List[i]; if( LeftBorderSum > MaxLeftBorderSum ) MaxLeftBorderSum = LeftBorderSum; } /* 左边扫描结束 */ MaxRightBorderSum = 0; RightBorderSum = 0; for( i=center+1; i<=right; i++ ) { /* 从中线向右扫描 */ RightBorderSum += List[i]; if( RightBorderSum > MaxRightBorderSum ) MaxRightBorderSum = RightBorderSum; } /* 右边扫描结束 */ /* 下面返回"治"的结果 */ return Max3( MaxLeftSum, MaxRightSum, MaxLeftBorderSum + MaxRightBorderSum ); } int MaxSubseqSum3( int List[], int N ) { /* 保持与前2种算法相同的函数接口 */ return DivideAndConquer( List, 0, N-1 ); }