最大子组问题

原创

躬行之zhi 2021-08-11 10:37:25 ©著作权

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#include "stdafx.h"
#include <stdio.h>

typedef struct Result
{
    Result():low(0), high(0), sum(0){}
    Result(int lowVal, int highVal, int sumVal) : low(lowVal)
    , high(highVal), sum(sumVal) {}
    int low;
    int high;
    int sum;
}Result;

Result findMaxCrossArray(int* arr, int low, int mid, int high)
{
    int crossleft = 0;
    int crossright = 0;
    int crossSum = 0;
    int leftSum = -0xffff;
    int sum = 0;
    for (int i = mid; i >= low; i--)
    {
        sum += arr[i];
        if (sum > leftSum)
        {
            leftSum = sum;
            crossleft = i;
        }
    }

    int rightsum = -0xffff;
    sum = 0;
    for (int j = mid + 1; j <= high; j++)
    {
        sum += arr[j];
        if (sum > leftSum)
        {
            rightsum = sum;
            crossright = j;
        }
    }

    return Result(crossleft, crossright, leftSum + rightsum);
}

Result findMaxMumSubArray(int* a, int low, int high)
{
    printf("called findMaxMumSubArray>> Low:%d, High:%d\n", low, high);
    Result* pResult = &Result();

    if (low == high)
    {
        pResult = &Result(low, high, a[low]);
        printf("oneItem------>low:%d, high:%d, sum:%d\n", pResult->low, pResult->high, pResult->sum);
    }
    else
    {
        int mid = (low + high) / 2;

        Result leftResult = findMaxMumSubArray(a, low, mid);
        printf("left------>low:%d, high:%d, sum:%d\n", leftResult.low, leftResult.high, leftResult.sum);
        Result rightResult = findMaxMumSubArray(a, mid + 1, high);
        printf("right------>low:%d, high:%d, sum:%d\n", rightResult.low, rightResult.high, rightResult.sum);
        Result crossResult = findMaxCrossArray(a, low, mid, high);
        printf("cross------>low:%d, high:%d, sum:%d\n", crossResult.low, crossResult.high, crossResult.sum);
        if (leftResult.sum >= rightResult.sum && leftResult.sum >= crossResult.sum)
        {
            pResult = &leftResult;
        }
        else if (rightResult.sum >= leftResult.sum && rightResult.sum >= crossResult.sum)
        {
            pResult = &rightResult;
        }
        else
            pResult = &crossResult;
    }

    printf(">>>>>>>>>result------>low:%d, high:%d, sum:%d\n", pResult->low, pResult->high, pResult->sum);
    return *pResult;

}

int _tmain(int argc, _TCHAR* argv[])
{
    int arr[] = { 13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7 };
    Result r = findMaxMumSubArray(arr, 0, 15);
    printf("low:%d, high:%d, sum:%d\n", r.low, r.high, r.sum);
    getchar();
   return 0;
}

-----------------------------------------------------------------------------

输出：

-------------------------------------------------------------------------------

called findMaxMumSubArray>> Low:0, High:15
called findMaxMumSubArray>> Low:0, High:7
called findMaxMumSubArray>> Low:0, High:3
called findMaxMumSubArray>> Low:0, High:1
called findMaxMumSubArray>> Low:0, High:0
oneItem------>low:0, high:0, sum:13
>>>>>>>>>result------>low:0, high:0, sum:13
left------>low:0, high:0, sum:13
called findMaxMumSubArray>> Low:1, High:1
oneItem------>low:1, high:1, sum:-3
>>>>>>>>>result------>low:1, high:1, sum:-3
right------>low:1, high:1, sum:-3
cross------>low:0, high:0, sum:-65522
>>>>>>>>>result------>low:0, high:0, sum:13
left------>low:0, high:0, sum:13
called findMaxMumSubArray>> Low:2, High:3
called findMaxMumSubArray>> Low:2, High:2
oneItem------>low:2, high:2, sum:-25
>>>>>>>>>result------>low:2, high:2, sum:-25
left------>low:2, high:2, sum:-25
called findMaxMumSubArray>> Low:3, High:3
oneItem------>low:3, high:3, sum:20
>>>>>>>>>result------>low:3, high:3, sum:20
right------>low:3, high:3, sum:20
cross------>low:2, high:3, sum:-5
>>>>>>>>>result------>low:3, high:3, sum:20
right------>low:3, high:3, sum:20
cross------>low:0, high:0, sum:-65525
>>>>>>>>>result------>low:3, high:3, sum:20
left------>low:3, high:3, sum:20
called findMaxMumSubArray>> Low:4, High:7
called findMaxMumSubArray>> Low:4, High:5
called findMaxMumSubArray>> Low:4, High:4
oneItem------>low:4, high:4, sum:-3
>>>>>>>>>result------>low:4, high:4, sum:-3
left------>low:4, high:4, sum:-3
called findMaxMumSubArray>> Low:5, High:5
oneItem------>low:5, high:5, sum:-16
>>>>>>>>>result------>low:5, high:5, sum:-16
right------>low:5, high:5, sum:-16
cross------>low:4, high:0, sum:-65538
>>>>>>>>>result------>low:4, high:4, sum:-3
left------>low:4, high:4, sum:-3
called findMaxMumSubArray>> Low:6, High:7
called findMaxMumSubArray>> Low:6, High:6
oneItem------>low:6, high:6, sum:-23
>>>>>>>>>result------>low:6, high:6, sum:-23
left------>low:6, high:6, sum:-23
called findMaxMumSubArray>> Low:7, High:7
oneItem------>low:7, high:7, sum:18
>>>>>>>>>result------>low:7, high:7, sum:18
right------>low:7, high:7, sum:18
cross------>low:6, high:7, sum:-5
>>>>>>>>>result------>low:7, high:7, sum:18
right------>low:7, high:7, sum:18
cross------>low:5, high:7, sum:-21
>>>>>>>>>result------>low:7, high:7, sum:18
right------>low:7, high:7, sum:18
cross------>low:3, high:0, sum:-65515
>>>>>>>>>result------>low:3, high:3, sum:20
left------>low:3, high:3, sum:20
called findMaxMumSubArray>> Low:8, High:15
called findMaxMumSubArray>> Low:8, High:11
called findMaxMumSubArray>> Low:8, High:9
called findMaxMumSubArray>> Low:8, High:8
oneItem------>low:8, high:8, sum:20
>>>>>>>>>result------>low:8, high:8, sum:20
left------>low:8, high:8, sum:20
called findMaxMumSubArray>> Low:9, High:9
oneItem------>low:9, high:9, sum:-7
>>>>>>>>>result------>low:9, high:9, sum:-7
right------>low:9, high:9, sum:-7
cross------>low:8, high:0, sum:-65515
>>>>>>>>>result------>low:8, high:8, sum:20
left------>low:8, high:8, sum:20
called findMaxMumSubArray>> Low:10, High:11
called findMaxMumSubArray>> Low:10, High:10
oneItem------>low:10, high:10, sum:12
>>>>>>>>>result------>low:10, high:10, sum:12
left------>low:10, high:10, sum:12
called findMaxMumSubArray>> Low:11, High:11
oneItem------>low:11, high:11, sum:-5
>>>>>>>>>result------>low:11, high:11, sum:-5
right------>low:11, high:11, sum:-5
cross------>low:10, high:0, sum:-65523
>>>>>>>>>result------>low:10, high:10, sum:12
right------>low:10, high:10, sum:12
cross------>low:8, high:0, sum:-65522
>>>>>>>>>result------>low:8, high:8, sum:20
left------>low:8, high:8, sum:20
called findMaxMumSubArray>> Low:12, High:15
called findMaxMumSubArray>> Low:12, High:13
called findMaxMumSubArray>> Low:12, High:12
oneItem------>low:12, high:12, sum:-22
>>>>>>>>>result------>low:12, high:12, sum:-22
left------>low:12, high:12, sum:-22
called findMaxMumSubArray>> Low:13, High:13
oneItem------>low:13, high:13, sum:15
>>>>>>>>>result------>low:13, high:13, sum:15
right------>low:13, high:13, sum:15
cross------>low:12, high:13, sum:-7
>>>>>>>>>result------>low:13, high:13, sum:15
left------>low:13, high:13, sum:15
called findMaxMumSubArray>> Low:14, High:15
called findMaxMumSubArray>> Low:14, High:14
oneItem------>low:14, high:14, sum:-4
>>>>>>>>>result------>low:14, high:14, sum:-4
left------>low:14, high:14, sum:-4
called findMaxMumSubArray>> Low:15, High:15
oneItem------>low:15, high:15, sum:7
>>>>>>>>>result------>low:15, high:15, sum:7
right------>low:15, high:15, sum:7
cross------>low:14, high:15, sum:3
>>>>>>>>>result------>low:15, high:15, sum:7
right------>low:15, high:15, sum:7
cross------>low:13, high:0, sum:-65520
>>>>>>>>>result------>low:13, high:13, sum:15
right------>low:13, high:13, sum:15
cross------>low:8, high:0, sum:-65515
>>>>>>>>>result------>low:8, high:8, sum:20
right------>low:8, high:8, sum:20
cross------>low:7, high:11, sum:38
>>>>>>>>>result------>low:7, high:11, sum:38
low:7, high:11, sum:38

-----------------------------------------------------------------------------------------

问题描述：
给定一只股票在某段时间内的历史价格变化曲线，找出一个能够实现收益最大化的时间段。
最大子组问题_3c
理解：
为找出最大化的收益，需要考虑的是在买进和卖出时的价格变化幅度，因此从该股票的每日变化幅度来考虑问题比较合适。由此，可以将上述问题稍作变形：给定一只股票在某段时间内的每日变化幅度，找出一个合适的买进和卖出时间，以实现收益最大化。因此，将输入数据转换如下，并试图在整个时间段中找到一个累加和最大的子区间，亦即最大子数组。

天	0	1	2	3	4
价格	10	11	7	10	6
变化		1	-4	3	-4

暴力求解方法：
        首先能够想到的是在一个给定数组（区间）中，其子数组（子区间）的个数是C(2,n)，很容易就能遍历完所有子数组从而找出最大的那个，其最坏情况渐进时间复杂度是Θ(n2)。假设每日变化幅度保存在数组A中（A的下标从1到n），A.length表示A的元素个数，最终结果以元组形式返回；给出伪码如下：
       BRUTE_FORCE(A)
           i = 1
           sum = -infinity
           for i <= A.length, inc by 1
               j = i
               last_sum = 0
               for j <= A.length, inc by 1
                   last_sum += A[j]
                   if last_sum > sum
                       sum = last_sum
                       start = i
                       end = j
           return (start, end, sum)

分治求解方法：
        上述方法的渐进时间复杂度差强人意。类比于归并排序，有时采用分治策略能够获得更好的时间复杂度。分治策略通常包含分解成子问题、解决子问题、合并子问题。由此可以推出大致的解决思路：首先依然假设数据输入如上一个方法那样，然后考虑将A[1...n]拆分为规模大致相同的两个子数组left[1...mid]和right[mid+1...n]，其中mid=(1+n)/2向下取整，那么可以肯定，最大子数组要么在这两个子数组中，要么横跨这两个子数组，因此可以分别求解这三种情况，取其中最大的子数组并返回即可。
        对于left/right子数组可递归求解，而对于横跨两个子数组的情况，如果能够使得该情况下的求解时间复杂度为O(n)，那么应该能让整体的最坏时间复杂度低于Θ(n2)。如果仅仅是通过遍历所有包含A[mid]和A[mid+1]的子数组来找最大子数组，那么很显然仅求解该情况就需要Θ(n2)的时间。可以推断横跨两个子数组的最大子数组，必须由两个分别在left/right中的子数组组成，这两个子数组在分别包含了A[mid]和A[mid+1]的所有子数组中是最大的；因为如果存在一个不满足上述条件的最大子数组，那么总可以用上述方法找到一个更大的子数组。
       根据上述思路，很容易推知求解横跨两个子数组的情况只需要O(n)的时间。由此给出伪码如下：
        （1）子过程：找出横跨两个子数组的最大子数组
           FIND_CROSSING_MAX_SUBARRAY(A, low, mid, high)
               left_sum = -infinity
               sum = 0
               i = mid
               for i >= low, dec by 1
                   sum += A[i]
                   if sum > left_sum
                       left_sum = sum
                       left_index = i

               right_sum = -infinity
               sum = 0
               i = mid + 1
               for i <= high, inc by 1
                   sum += A[i]
                   if sum > right_sum
                       right_sum = sum
                       right_index = i
               return (left_index, right_index, left_sum+right_sum)

        （2）主过程：分治法找出最大子数组
           FIND_MAX_SUBARRAY(A, low, high)
               if low == high
                   return (low, high, A[low])
               else
                   mid = down_trunc((low + high) / 2)
                   (left_start, left_end, left_sum) =
                       FIND_MAX_SUBARRAY(A, low, mid)
                   (right_start, right_end, right_sum) =
                       FIND_MAX_SUBARRAY(A, mid+1, high)
                   (cross_start, cross_end, cross_sum) =
                       FIND_CROSSING_MAX_SUBARRAY(A, low, mid, high)

                   if left_sum > right_sum and left_sum > cross_sum
                       return (left_start, left_end, left_sum)
                   else if right_sum > left_sum and right_sum > cross_sum
                       return (right_start, right_end, right_sum)
                   else
                       return (cross_start, cross_end, cross_sum)
        可以看出上述算法渐进时间复杂度为Θ(nlg(n))。

缩减问题规模的方法：
在查找过程中，是否可以根据现有的信息，来缩减需要排查的子数组个数，进而获得更好的时间复杂度呢？一个思路是不再重复检查以前累加过的元素，即从左至右累加元素，保存其中的最大子数组，如果在加入一个元素后累加和为负数，则从该元素的后一个元素重新累加，直至整个数组遍历完毕。该思路有效的前提是证明以下几个假设：

可以将最大子数组来源分为三种：已经遍历完的数组部分、未遍历的数组部分以及跨越这两部分的子数组
可以假设当从左至右累加直至累加和为负，所得的最大子数组是当前已遍历完的数组部分中最大的
可以假设当累加和为负时，潜在的最大子数组不可能从该元素或该元素左边的元素开始

        假设1不证自明。
       假设从A[1]累加到A[i]时第一次遇到其累加和为负（1<=i<=n），那么A[i]一定为负，且A[1]+...+A[i-1]>=0。当i<=2时，显然此时假设2成立。当i>2时，可以认为在A[1]...A[i]中，所有子数组可分为三种：从A[1]开始向右拓展、从A[i]开始向左拓展以及不包含A[1]和A[i]的中间子数组；显然从A[i]向左拓展的不可能是最大子数组，而如果不包含A[1]和A[i]的中间子数组是最大子数组，那么可以使该中间子数组加上其左边的部分构成一个新的子数组，而且该子数组总是大于等于这个中间子数组，因为其左边部分总是大于等于0，所以该情况下假设2也得证。综合来看假设2是成立的。
        对于假设3，显然潜在的最大子数组不可能从A[i]开始，因为A[i]<0。当潜在的最大子数组从A[i]的左边开始时，假设其从A[j]开始（1<=j<i）。显然j不能等于1，因为A[1]+...+A[i]<0；当j>1时，A[j]+...+A[i]一定是负数，因为A[1]+...+A[j-1]一定大于等于0而A[1]+...+A[i]一定为负。所以综合来看，从A[i]或者A[i]的左边寻找潜在的子数组是没有意义的。
        伪码如下，时间复杂度为Θ(n)。对于全部是负数的情况，特殊处理即可，不影响时间复杂度。
       LINEAR_SEARCH_MAX_SUBARRAY(A)
           sum = -infinity
           start = 0
           end = 0

           cur_sum = 0
           cur_start_index = 1

           i = 1
           for i <= A.length, inc by 1
               cur_sum += A[i]
               if cur_sum < 0
                   cur_sum = 0
                   cur_start_index = i + 1
               else
                   if sum < cur_sum
                       sum = cur_sum
                       start = cur_start_index
                       end = i

           return (start, end, sum)