分治策略 | 最大子数组问题

分治策略  就是把一个大问题递归地分解成规模较小的子问题，子问题的规模可能不等。

最大子数组问题，找数组A中和最大的连续段A[i, .., j]。A中要有负数这个问题才有意义。





伪代码：

（跨中点的情况）



 


（完全在原数组左边或右边的情况）

  


C++代码：

#include <iostream>

using namespace std;

struct maxSubarray
{
int low;
int high;
int sum;
};

maxSubarray findMaxCrossingSubarray(int* arr, int low, int mid, int high)
{
int sum = 0;
int leftSum = -65535;
int rightSum = -65535;
maxSubarray crossSub;
crossSub.low = low;
crossSub.high = high;
for (int i = mid; i >= low; i--)
{
sum += arr[i];
if (sum > leftSum)
{
leftSum = sum;
crossSub.low = i;
}
}
sum = 0;
for (int i = mid + 1; i <= high; i++)
{
sum += arr[i];
if (sum > rightSum)
{
rightSum = sum;
crossSub.high = i;
}
}
crossSub.sum = leftSum + rightSum;
return crossSub;
}

maxSubarray findMaxmumSubarray(int* arr, int low, int high, int sum)
{
if (low == high)
{
maxSubarray sub;
sub.low = low;
sub.high = high;
sub.sum = arr[low];
return sub;
}
else
{
int mid = (low + high) / 2;
maxSubarray leftSub;
leftSub = findMaxmumSubarray(arr, low, mid, sum);
maxSubarray rightSub;
rightSub = findMaxmumSubarray(arr, mid + 1, high, sum);
maxSubarray crossSub;
crossSub = findMaxCrossingSubarray(arr, low, mid, high);
if (leftSub.sum >= rightSub.sum && leftSub.sum >= crossSub.sum)
return leftSub;
else if (rightSub.sum >= leftSub.sum && rightSub.sum >= crossSub.sum)
return rightSub;
else if (crossSub.sum >= leftSub.sum && crossSub.sum >= rightSub.sum)
return crossSub;
}
}

int main()
{
int A[10] = { 4, 5, 7, -2, -3, -6, -1, 9, -2, 4 };
maxSubarray maxSub;
maxSub = findMaxmumSubarray(A, 0, sizeof(A) / sizeof(A[0]) - 1, -65535);
cout << maxSub.low << "," << maxSub.high << "," << maxSub.sum << endl;
return 0;
}

---------- 更新分治策略写法 (上面的代码是何等的woc...) ----------

int maxSubArray(vector<int>& nums) {
if (nums.empty())
return 0;
return findMaxSubArray(nums, 0, nums.size() - 1);
}

int findMaxSubArray(vector<int>& nums, int left, int right) {
if (left >= right) return nums[left];
int mid = left + (right - left) / 2;
int lmax = findMaxSubArray(nums, left, mid - 1);
int rmax = findMaxSubArray(nums, mid + 1, right);
int mmax = nums[mid], temp = mmax; //跨中点数组
for (int i = mid - 1; i >= left; --i) {
temp += nums[i];
mmax = max(mmax, temp);
}
temp = mmax;
for (int i = mid + 1; i <= right; ++i) {
temp += nums[i];
mmax = max(mmax, temp);
}
return max(mmax, max(lmax, rmax));
}

习题中也介绍了一种方法，线性时间O(n)指的是已知A[1,..,j]的最大子数组再求A[1,..,j+1]的最大子数组，但如果看整体的时间复杂度，还是O(n^2)，不如分治法好。