求非连续子序列的问题用DP很好解决,但是**分治法**就比较难了
假设有一个序列是L = [1, 0, 5, 3, 2, 7, 9, 15, 6, 4, 13]
他的最大非连续子序列就是 S = [1, 5, 7, 15, 13] 俩俩数字任意不相邻
现在要求给L求S
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MAX_N 20
int max3(int, int, int);
int maxSubArrayAns(int []);
int maxSubArray(int [], int, int);
int main(){
int nums[MAX_N];
int i;
srand(time(0));
printf("array: \n");
for(int i = 0; i < MAX_N; i++){
nums[i] = (int)(rand() % (MAX_N * 2) - MAX_N);
printf("%d\t", nums[i]);
}
printf("\n");
printf("The max subsequen sum is %d.\n", maxSubArrayAns(nums));
return 0;
}
int max3(int a, int b, int c){
if(a > b)
return a > c ? a : c;
else
return b > c ? b : c;
}
int maxSubArray(int nums[], int left, int right){
int maxLeftSum, maxRightSum;
int maxLeftBorderSum, maxRightBorderSum;
int leftBorderSum, rightBorderSum;
if(left == right)
if(nums[left] > 0)
return nums[left];
else
return 0;
int mid = (left + right) / 2, i;
maxLeftSum = maxSubArray(nums, left, mid);
maxRightSum = maxSubArray(nums, mid + 1, right);
maxLeftBorderSum = 0, leftBorderSum = 0;
for(i = mid; i >= left; i--){
leftBorderSum += nums[i];
if(leftBorderSum > maxLeftBorderSum)
maxLeftBorderSum = leftBorderSum;
}
maxRightBorderSum = 0, rightBorderSum = 0;
for(i = mid + 1; i <= right; i++){
rightBorderSum += nums[i];
if(rightBorderSum > maxRightBorderSum)
maxRightBorderSum = rightBorderSum;
}
return max3(maxLeftSum, maxRightSum, maxLeftBorderSum + maxRightBorderSum);
}
int maxSubArrayAns(int nums[]){
return maxSubArray(nums, 0, MAX_N - 1);
}
使用分治法的话,平均时间复杂度为Θ(n lg n)。实际上解决最大子序列问题还有一种更加快速的方法,这种方法的时间复杂度是Θ(n),是一种线性的算法
int maxSubArrayAns(int nums[]){
int i, thisSum = 0, maxSum = 0;
for(i = 0; i < MAX_N - 1; i++){
thisSum += nums[i];
if(thisSum > maxSum)
maxSum = thisSum;
else if(thisSum < 0)
thisSum = 0;
}
return maxSum;
}