leetCode 53. maximum subarray

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.

 

贪婪算法找每个当前位置对应的最大的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;}
};

　　

更简单的方法，记录当前最大的结果，因为每遇到一个数，可能有两种可能，一是加这个数，二是从这个数开始。 那我们就要找当前的最优情况与历史最优情况比较。

和可能的结果：

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);//记录加当前数与从当前数开始的最大值cont_sum = max(cont_sum, adj_sum);// 比较当前的最大值与历史最大值，记录最大值}return cont_sum;}
};

　　

 

这是一个最优化问题，最优化问题一般都可以用DP（动态规划）解决。 对于动态规划，要考虑子问题是什么（形式的子问题或状态的子问题）子问题就可以用recursive solution。

1. 对于 maxSubArray( int a[], int i,int j), is searching for the maxSubArray for a[i:j]
2. the goal is to figure out what maxSubArray(A,0,A.length()-1) is.
3. 对于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; }

　　开始写 dp[i]=nums[i]+dp[i-1]>0?dp[i-1]:0;  有错哈哈， 忘了带括号

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