代码实现如下: 1: int maxSubArray(int A[], int n) {2: // Start typing your C/C++ solution below3: // DO NOT write int main() function4: int maxV = INT_MIN;5: return maxArray(A, 0, n-1, maxV);6: }7: int maxArray(int A[], int left, int right, int& maxV)8: {9...
此题的子问题可以定为 maxSubArray(vector<int> &nums, int i); 就是在子序列nums[0:i] 中的最大序列。 classSolution {public:intmaxSubArray(vector<int>&nums) {intn =nums.size();if(n ==1)returnnums[0];//if (n == 2) return max(nums[0],nums[1]);vector<int>dp(n); dp[0] = ...
Given an array of integers, find a contiguous subarray which has the largest sum. Notice:The subarray should contain at least one number. 给定一个整数数组,找到一个具有最大和的子数组,返回其最大和。 注意:子数组最少包含一个数 【题目链接】Maximum Subarray 【题目解析】 O(n)就是一维DP. 假设A...
Given an array of integers, find a contiguous subarray which has the largest sum. Notice:The subarray should contain at least one number. 给定一个整数数组,找到一个具有最大和的子数组,返回其最大和。 注意:子数组最少包含一个数 【题目链接】 http://www.lintcode.com/en/problem/maximum-subarray/...
【数据结构】算法 Maximum Subarray 最大子数组:Maximum Subarray 参考来源:Maximum subarray problem Kadane算法扫描一次整个数列的所有数值,在每一个扫描点计算以该点数值为结束点的子数列的最大和(正数和)。该子数列由两部分组成:以前一个位置为结束点的最大子数列、该位置的数值。因为该算法用到了“最佳子结构”...
http://www.lintcode.com/en/problem/maximum-subarray-iii/ 【题目解析】 最重要的思路: 维护一个 globalMax [ k+1 ] [ len+1 ] 的矩阵, globalMax [ i ] [ j ] 代表了前 j 个数中取 i 个 subarray 得到的最大和, 注意这里第 j 个数不一定被包括。
[LeetCode&Python] Problem 53. Maximum Subarray Given an integer arraynums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. Example: Input: [-2,1,-3,4,-1,2,1,-5,4],
1classSolution{2public:3intmaxSubArray(intA[],intn){4intres, lres, rres, all;5maxSubArray(A,0, n -1, res, lres, rres, all);6returnres;7}8private:9voidmaxSubArray(intA[],intl,intr,int&res,int&lres,int&rres,int&all){10if(l ==r)11{12res = lres = rres = all =A[l...
http://en.wikipedia.org/wiki/Maximum_subarray_problem 并发现,这道题有两种经典解法,一个是:Kadane算法,算法复杂度O(n);另外一个是分治法:算法复杂度为O(nlogn)。 1. Kadane算法 代码如下: 1publicintmaxSubArray(int[] A) { 2intmax_ending_here = 0; ...
problem... The idea is to save the answer for a suffix and a prefix subaaray for particular segment and also saving the subarray with maximum( sum* length) for that segemnt. Then keep merging them to get the answer for the whole array. I implemented a solution using segment trees (...