53. Maximum Subarray
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Problem:
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.
Code1:(超时)
class Solution {public: int maxSubArray(vector<int>& nums) { int sum = 0, large = nums[0]; for (int i = 0; i < nums.size(); i++) { sum = 0; for (int j = i; j < nums.size(); j++) { sum += nums[j]; if (sum > large) { large = sum; } } } return large; }};
一开始想着随便写个思路比较简单的算法,然后发现这个做法是超时了= =。这个复杂度是O(n²),所以超时也是很正常的,一旦n无限大,肯定是超时的。然后就换了个做法。
Code2:
class Solution {public: int maxSubArray(vector<int>& nums) { int largest = nums[0]; int sum = 0; for (int i = 0; i < nums.size(); i++) { sum += nums[i]; if (sum > largest) { largest = sum; } //负数+下一项永远小于下一项,所以sum置0,重新开始比较。 if (sum < 0) { sum = 0; } } return largest; }};
这个做法是参考了别人写的,觉得很厉害,也很高效。主要思想是只要和小于0的时候,sum就置0,因为负数加下一项,永远是下一项更大的,所以可以置0重新开始计算,也不会影响到开始得到的largest的结果。这个算法的复杂度是O(n)。
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