LeetCode(53) Maximum Subarray
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.
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
最大子序列和的问题,这道题我写出的是O(n)的算法,属于简单的动态规划,根据题目后面的more practice说明该题目还有更优的分治法解决思路。
AC代码-动态规划
class Solution {
public:
int maxSubArray(vector<int>& nums) {
if (nums.empty())
return 0;
//求数组的长度
int len = nums.size();
//将最大和赋值为首元素值,temp记录临时子序列和
int maxSum = nums[0], temp = 0;
for (int i = 0; i < len; i++)
{
temp += nums[i];
//若元素和大于当前最大和
if(temp > maxSum)
{
maxSum = temp;
}//else
//若子系列和为非正数,则从下一个元素重新记录
if (temp <= 0)
{
temp = 0;
}
}//for
return maxSum;
}
};
AC代码-分治法
class Solution {
public:
int maxSubArray(vector<int>& nums) {
if (nums.empty())
return 0;
//求数组的长度
int len = nums.size();
return Divide(nums , 0 , len-1);
}
//分治法
int Divide(const vector<int> &nums, int lhs, int rhs)
{
if (lhs == rhs)
return nums[lhs];
int mid = (lhs + rhs) / 2;
int leftMaxSum = Divide(nums, lhs, mid);
int rightMaxSum = Divide(nums, mid + 1, rhs);
int lsum = INT_MIN;
int rsum = INT_MIN;
int temp = 0;
for (int i = mid; i >= lhs; i--)
{
temp += nums[i];
if (temp > lsum)
lsum = temp;
}
temp = 0;
for (int i = mid + 1; i <= rhs; i++)
{
temp += nums[i];
if (temp > rsum)
rsum = temp;
}
//跨越中点的最大子序列和
temp = lsum + rsum;
return std::max(temp, std::max(leftMaxSum, rightMaxSum));
}
};
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