《编程之美》- 2.14 - 求数组的子数组之和的最大值

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题目

一个有N个整数元素的一维数组，{A[0]...A[n-1]}，求该数组所有子数组中的最大和。

题意明确：

题目中说的子数组是连续的；
题目只需要求和，并不需要返回子数组的具体位置；
数组的元素是整数，所以数组可能包含有正整数，零和负整数；

例：

数组[ 1,-2,3,5,-3,2 ]返回8；

分析

《编程之美》P183-188中详细讲述了，该题目从最初暴力O(n^3)的解法，优化到最后O(N)的步骤；

代码

/*2.14 求数组的子数组之和的最大值*/#include <iostream>#include <cstdlib>#include <vector>#include <algorithm>using namespace std;/*方法一：罗列所有的子数组，求每个子数组的和，时间复杂度O(n^3)*/int maxSubSum(const vector<int> &nums){if (nums.empty()){return 0;}//ifint len = nums.size();int maxSum = INT_MIN, tmpSum = 0;for (int i = 0; i < len; ++i){for (int j = i; j < len; ++j){/*清0临时子数组的和*/tmpSum = 0;/*求i到j的子数组之和*/for (int k = i; k <= j; ++k){tmpSum += nums[k];}//forif (tmpSum > maxSum){maxSum = tmpSum;}//if}//for}//forreturn maxSum;}/*方法二：注意到sum[i...j] = sum[i-1 ... j] + nums[j] , 复杂度降低到O(n^2)*/int maxSubSum2(const vector<int> &nums){if (nums.empty()){return 0;}//ifint len = nums.size();int maxSum = INT_MIN, tmpSum = 0;for (int i = 0; i < len; ++i){tmpSum = 0;for (int j = i; j < len; ++j){tmpSum += nums[j];/*从i到j的子数组之和为tmpSum*/if (tmpSum > maxSum)maxSum = tmpSum;}//for}//forreturn maxSum;}/*方法三：分治法的思想，分为规模减半的子问题，通过递归解决；复杂度O(nlogn)*//*方法四：动态规划max{A[0], A[0]+start[1], ALL[1]}，时间复杂度O(N)，空间复杂度O(N)*/int maxSubSum4(const vector<int> &A){if (A.empty()){return 0;}//ifint len = A.size();/*Start数组中存放的是，以下标i对应元素为起始点的最大子数组之和*/vector<int> Start(len, 0);/*All数组中存放的是，从下标[i...len-1]实际最大子数组之和*/vector<int> All(len, 0);/*从数组末尾向前遍历，知道数组首部*/Start[len - 1] = A[len - 1];All[len - 1] = A[len - 1];for (int i = len - 2; i >= 0; --i){Start[i] = max(A[i], A[i] + Start[i + 1]);All [i] = max(Start[i], All[i + 1]);}//forreturn All[0];}/*方法五：四的改进，空间复杂度减低到O(1)*/int maxSubSum5(const vector<int> &A){if (A.empty()){return 0;}//ifint len = A.size();int nStart = A[len - 1], nAll = A[len - 1];for (int i = len - 2; i >= 0; --i){nStart = max(A[i], nStart + A[i]);nAll = max(nStart, nAll);}//forreturn nAll;}/*方法六：线性时间复杂度*/int maxSubSum6(const vector<int> &A){if (A.empty()){return 0;}//if/*求数组A的长度*/int n = A.size();int maxSum = A[n - 1], tmpSum = A[n - 1];for (int i = n - 2; i >= 0; --i){if (tmpSum < 0){tmpSum = 0;}//iftmpSum += A[i];if (tmpSum > maxSum)maxSum = tmpSum;}//forreturn maxSum;}int main(){vector<int> A = { 1,-2,3,5,-3,2 };cout << maxSubSum(A) << endl;cout << maxSubSum2(A) << endl;cout << maxSubSum4(A) << endl;cout << maxSubSum5(A) << endl;cout << maxSubSum6(A) << endl;system("pause");return 0;}

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