leecode 解题总结：53. Maximum Subarray

#include <iostream>#include <stdio.h>#include <vector>using namespace std;/*问题:Find the contiguous subarray within an array (containing at least one number) which has the largest sum.For example, given the array [-2,1,-3,4,-1,2,1,-5,4],the contiguous subarray [4,-1,2,1] has the largest sum = 6.click to show more practice.More practice:If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.分析:最大连续子数组问题。可以用动态规划来做。设dp[i]表示以元素A[i]结尾的最大连续子数组的和，则有dp[i+1] = {A[i+1] ,if dp[i] < 0          {dp[i] + A[i] , else 最后返回dp数组中最大值即可输入:9-2 1 -3 4 -1 2 1 -5 4输出:6*/class Solution {public:    int maxSubArray(vector<int>& nums) {if(nums.empty()){return 0;}int size = nums.size();        vector<int> dp(size , 0);dp.at(0) = nums.at(0);int maxValue = dp.at(0) ;for(int i = 1 ; i < size ; i++){if(dp.at(i-1) < 0){dp.at(i) = nums.at(i); }else{dp.at(i) = dp.at(i-1) + nums.at(i);}if(dp.at(i) > maxValue){maxValue = dp.at(i);}}return maxValue;    }};void process(){int num;int value;vector<int> nums;Solution solution;while(cin >> num){nums.clear();for(int i = 0 ; i < num ; i++){cin >> value;nums.push_back(value);}int result = solution.maxSubArray(nums);cout << result << endl;}}int main(int argc , char* argv[]){process();getchar();return 0;}