标签：etc ++ max nbsp public bar number 问题 连续子序列
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.
题解：
由于是连续子序列这个限制，所以如果k+1这个元素之前的和是小于0的，那么对于增大k+1这个元素从而去组成最大子序列是没有贡献的，所以可以把sum置0处理。
本质上是DP问题。，只是要先求出sum。
Solution 1 ()
class Solution { public: int maxSubArray(vector<int>& nums) { int res = INT_MIN, sum = 0; for (int n : nums) { /* if(sum < 0) sum = 0; sum += A[i]; res = max(res, sum); */ sum = max(sum + n, n); res = max(res, sum); } return cur; } };
Solution 2 (TLE)
class Solution { public: int helper(vector<int> nums, int left, int right) { if(left >= right) return nums[left]; int mid = left + (right - left)/2; int lmax = helper(nums, left, mid-1); int rmax = helper(nums, mid+1, right); int mmax = nums[mid], tmp = mmax; for(int i=mid-1; i>=left; --i) { tmp += nums[i]; mmax = max(tmp, mmax); } tmp = mmax; for(int i=mid+1; i<=right; ++i) { tmp += nums[i]; mmax = max(tmp, mmax); } return max(mmax, max(lmax, rmax)); } int maxSubArray(vector<int>& nums) { if(nums.size() == 0) return 0; return helper(nums,0, nums.size()-1); } };
【LeetCode】053. Maximum Subarray
标签：etc ++ max nbsp public bar number 问题 连续子序列
原文地址：http://www.cnblogs.com/Atanisi/p/6816196.html
