LeetCode 53. Maximum Subarray

最新推荐文章于 2022-10-11 20:23:47 发布

SYSUJackJiao

最新推荐文章于 2022-10-11 20:23:47 发布

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本文探讨了寻找具有最大和的连续子数组的问题，并提供了两种不同的解决方案：暴力解法和分治法。暴力解法虽然直观但效率较低，时间复杂度为O(n^2)；而分治法则更为高效，其时间复杂度为O(nlogn)。

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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.

解法一：暴力解法

#include<vector>
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int max = -99999;
for (vector<int>::iterator i = nums.begin() ; i!=nums.end(); i++) {
int sum = 0;
for (vector<int>::iterator j = i; j != nums.end(); j++) {
sum += *j;
if (sum > max) max = sum;
}
}
return max;
}
};

这种解法属于从开始遍历，时间复杂度为O(n^2)，LeetCode上通过了所有的测试用例，但是属于Time Limit Exceeded。

解法二：分治法

#include<vector>
class Solution {
public:
int find_max (int a , int b , int c) {
int max = a;
if (b > a) max = b;
if (c > max) max = c;
return max;
}
int divided_maxSubArray(vector<int>& nums , int left , int right) {
if (left == right) return nums[left];
int mid = (left + right) / 2;

int leftmax = nums[mid];
int leftsum = 0;
for (int i = mid ; i >= left ; i--) {
leftsum += nums[i];
if (leftsum > leftmax) leftmax = leftsum;
}

int rightmax = nums[mid + 1];
int rightsum = 0;
for (int j = mid + 1 ; j<= right ; j++) {
rightsum += nums[j];
if (rightsum > rightmax) rightmax = rightsum;
}

return find_max(leftmax + rightmax , divided_maxSubArray(nums , left , mid) , divided_maxSubArray(nums, mid + 1, right));
}
int maxSubArray(vector<int>& nums) {
return divided_maxSubArray(nums , 0 , nums.size() - 1);
}
};

时间复杂度为O(nlogn)