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
.
查找给定数组中连续子数组最大的和。连续相加数组中的元素知道和小于0,如果这个和为负数,就让sum = 0,相当于抛弃了前面的子数组。
class Solution { public: int maxSubArray(vector<int>& nums) { int res = INT_MIN, sum = 0; for (int num : nums) { sum += num; res = max(res, sum); sum = max(sum, 0); } return res; } }; // 9 ms
这是一个最优化问题,所以可以使用动态规划的方法求解,首先需要找出子问题关系表达式,dp数组表示最大子数组的结束。dp[i] = nums[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0).
class Solution { public: int maxSubArray(vector<int>& nums) { vector<int> dp(nums.size(), nums[0]); int res = dp[0]; for (int i = 1; i != nums.size(); i++) { dp[i] = nums[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0); res = max(res, dp[i]); } return res; } }; // 12 ms