题目：
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.
More practice:
思路：
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
这道题目是个简单的动态规划问题，可以建立一个数组来存储每次的局部最优结果，然后再用循环比较求出最大的和
代码：
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int num = nums.size();
int maxSum[num];
int i;
maxSum[0] = nums[0];
for(i=1;i<num;i++)
{
maxSum[i] =max(maxSum[i-1]+nums[i],nums[i]);
}
for(i=1;i<num;i++)
{
if(maxSum[0]<maxSum[i])
maxSum[0] = maxSum[i];
}
return maxSum[0];
}
};
/*
time limit //本质上下面两个算法都是贪心算法
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int num = nums.size();
int sum[num] ;
int temp =0;
int i = num-1;
if(num==1)
return nums[0];
for( i ; i>= 0; --i )
{
sum[i] = accumulate(nums.begin(),nums.begin()+i+1, 0);
// cout<<sum[i]<<endl;
if(i==num-1)
continue;
temp = sum[i];
for(int j = 0;j<num-i-1;j++)
{
temp = nums[j+i+1]+temp-nums[j];
if(sum[i]<temp)
sum[i] = temp;
}
}
for (i=0;i<num-1;i++)
{
cout<<sum[i]<<endl;
if(sum[0]<sum[i+1])
sum[0] = sum[i+1];
}
return sum[0];
}
};
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int num = nums.size();
int sum[num] ;
int temp;
int i = num-1;
if(num==1)
return nums[0];
for( i ; i>= 0; --i )
{
sum[i] = accumulate(nums.begin(),nums.begin()+i+1, 0);
// cout<<sum[i]<<endl;
if(i==num-1)
continue;
for(int j = 0;j<num-i;j++)
{
int temp = accumulate(nums.begin()+j,nums.begin()+j+i+1, 0);
if(sum[i]< temp )
sum[i] =temp;
}
}
for (i=0;i<num-1;i++)
{
cout<<sum[i]<<endl;
if(sum[0]<sum[i+1])
sum[0] = sum[i+1];
}
return sum[0];
}
};
*/
