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.

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貪婪算法找每個當前位置對應的最大的subarray,?

class Solution { public:int maxSubArray(vector<int>& nums) {int n=nums.size();vector<int> eachnum;int sum=0;for(int i=0;i<n;i++){if(sum<0){sum=0;}sum+=nums[i];eachnum.push_back(sum);cout<<sum<<endl;// eachnum[i]=sum;}vector<int>:: iterator biggest=max_element(eachnum.begin(),eachnum.end());return *biggest;} };

　　

更簡單的方法，記錄當前最大的結果，因為每遇到一個數，可能有兩種可能，一是加這個數，二是從這個數開始。 那我們就要找當前的最優(yōu)情況與歷史最優(yōu)情況比較。

和可能的結果：

class Solution { public:int maxSubArray(vector<int>& nums) {int adj_sum = 0;int cont_sum = nums[0];for (vector<int>::iterator it = nums.begin(); it<nums.end(); it++){adj_sum+=*it;adj_sum = max(adj_sum, *it);//記錄加當前數與從當前數開始的最大值cont_sum = max(cont_sum, adj_sum);// 比較當前的最大值與歷史最大值，記錄最大值}return cont_sum;} };

　　

?

這是一個最優(yōu)化問題，最優(yōu)化問題一般都可以用DP（動態(tài)規(guī)劃）解決。 對于動態(tài)規(guī)劃，要考慮子問題是什么（形式的子問題或狀態(tài)的子問題）子問題就可以用recursive solution。

對于 maxSubArray( int a[], int i,int j), is searching for the maxSubArray for a[i:j]

the goal is to figure out what maxSubArray(A,0,A.length()-1) is.

對于maxSubArray(int a[], int i, int j) is difficult ot connect this sub problem to the original, so we change the format of the sub problem to maxSubArray(int a[], int i), which means the maxSubArray for A[0:i]. which must has A[i] as the end. so we should keep track of each solution of the sub problem to update the global optimal value.

maxSubArray(A,i)=maxSubArray(A, i-1)>0?maxSubArray(A, i-1):0+A[i];

so the solution is same as the previous one:

int maxSubArray(vector<int> & nums){int n=nums.size();int num=0;int* dp=new int[n];dp[0]=nums[0];int maxsum=nums[0];for (int i=1;i<n;i++){if(dp[i-1]<0)dp[i]=nums[i];elsedp[i]=dp[i-1]+nums[i];cout<<dp[i]<<" "<<dp[i-1]<<endl;maxsum=max(maxsum,dp[i]);cout<<maxsum<<endl;}return maxsum; }

　　開始寫 dp[i]=nums[i]+dp[i-1]>0?dp[i-1]:0; ?有錯哈哈， 忘了帶括號

? ? ? ? ? ? ? ? ? ?dp[i]=nums[i]+(dp[i-1]>0?dp[i-1]:0);?

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轉載于:https://www.cnblogs.com/fanhaha/p/7222002.html

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