Leetcode题解 53. Maximum Subarray 思路解析
2017-08-17 18:12
399 查看
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.
思路解析
就我的个人理解,这是一道即可看似动态规划,又可看似贪心问题的题目。在遍历的每次迭代中,我们维护两个变量,一个是全局最优global,就是到当前元素为止最优的解,一个是局部最优local,就是必须包含当前元素的最优的解。
(1)对于local,如果加上当前值后,变成了负数,说明了2个事实:
i:当前值是负数
ii:此时的local已经没有价值了,如果后续的连续子序列还包含目前这个子序列,那么肯定不如不包含当前这个子序列的sum大
所以这时候直接把local重置为0,连续子序列重新开始生成。
(2)对于global,每次迭代都与当前local比较,然后取二者中的最大值
整个过程只需要遍历完一次原数组,所以时间复杂度是O(n)
C++实现
// Runtime: 9 ms class Solution { public: int maxSubArray(vector<int>& nums) { if (0 == nums.size()) return 0; int global = nums[0], local = 0; for (int i = 0; i < nums.size(); ++i){ local += nums[i]; global = max(local, global); local = max(0, local); } return global; } };
Python实现
class Solution(object): def maxSubArray(self, nums): """ :type nums: List[int] :rtype: int :Runtime: 42 ms """ if 0 == len(nums): return 0 global_sum = nums[0] local_sum = 0 for i in nums: local_sum += i global_sum = max(local_sum, global_sum) local_sum = max(0, local_sum) return global_sum
相关文章推荐
- leetcode题解-53. Maximum Subarray && 448. Find All Numbers Disappeared in an Array
- Leetcode题解 - 53. Maximum Subarray
- [LeetCode] 53. Maximum Subarray 解题思路
- 【Leetcode题解】53. Maximum Subarray
- leetcode题解——53. Maximum Subarray
- LeetCode53. Maximum Subarray
- leetcode-53. Maximum Subarray
- LeetCode 53. Maximum Subarray
- Leetcode 300. Longest Increasing Subsequences (nlogn复杂度)思路解析
- 【LeetCode】53. Maximum Subarray
- LeetCode||53. Maximum Subarray
- 53. Maximum Subarray LeetCode
- [LeetCode] Algorithms-53. Maximum Subarray
- LeetCode 53. Maximum Subarray
- [array] leetcode - 53. Maximum Subarray - Easy
- leetcode 53. Maximum Subarray-最大子数组|动态规划
- leetcode 53. Maximum Subarray python
- 【Leetcode】53. Maximum Subarray
- [LeetCode]题解(python):152-Maximum Product Subarray
- leetcode--3,Longest Substring Without Repeating Characters &&53. Maximum Subarray&&String.valueOf()