思路
方法一：动态规划
1 class Solution { 2 public: 3 int maxSubArray(vector<int>& nums) { 4 //memo[i]表示以nums[i]结尾并且包含nums[i]的子数组的最大和 5 vector<int> memo(nums.size()); 6 memo[0] = nums[0]; 7 8 for(int i = 1; i < nums.size(); ++i) { 9 if(memo[i-1] > 0) 10 memo[i] = memo[i-1] + nums[i]; 11 else 12 memo[i] = nums[i]; 13 } 14 15 int maxSum = memo[0]; 16 for(int i = 1; i < memo.size(); ++i) { 17 if(memo[i] > maxSum) 18 maxSum = memo[i]; 19 } 20 21 return maxSum; 22 } 23 };
复杂度分析
时间复杂度：O(n)
空间复杂度：O(n)
方法二：模拟
1 class Solution { 2 public: 3 int maxSubArray(vector<int>& nums) { 4 int maxSum = nums[0]; 5 int thisSum = nums[0]; 6 7 for(int i = 1; i < nums.size(); ++i) { 8 if(thisSum < 0) 9 thisSum = 0; 10 11 thisSum += nums[i]; 12 if(thisSum > maxSum) 13 maxSum = thisSum; 14 } 15 16 return maxSum; 17 } 18 };
复杂度分析
时间复杂度：O(n)
空间复杂度：O(1)