leetcode-----53. 最大子序和

• 链接：https://leetcode-cn.com/problems/maximum-subarray/

算法1（前缀和暴力法）

时间复杂度：\(O(n^2)\)

代码

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int res = INT_MIN, sum = 0;
        for (int i = 0; i < nums.size(); ++i) {
            sum = 0;
            for (int j = i; j < nums.size(); ++j) {
                sum += nums[j];
                res = max(res, sum);
            }
        }
        return res;
    }
};

算法2（前缀和优化）

时间复杂度：\(O(n)\)

代码

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int res = nums[0], sum = 0, min_sum = 0;
        for (auto x: nums) {
            sum += x;
            res = max(res, sum - min_sum);
            min_sum = min(min_sum, sum);
        }
    return res;
    }
};

算法3（归并法）

时间复杂度：\(O(nlogn)\)

代码

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        if (nums.size() == 0) return 0;
        return help(nums, 0, nums.size() -1);
    }

    int help(vector<int> nums, int l, int r) {
        if (l > r) return INT_MIN;
        int mid = (l + r) >> 1;
        int left = help(nums, l, mid - 1);
        int right = help(nums, mid + 1, r);
        int l_sum = 0;
        int sum = 0;
        for (int i = mid - 1; i >= l; --i) {
            sum += nums[i];
            l_sum = max(l_sum, sum);
        }
        int r_sum = 0;
        sum = 0;
        for (int i = mid + 1; i <= r; ++i) {
            sum += nums[i];
            r_sum = max(sum, r_sum);
        }
        return max(l_sum + r_sum + nums[mid], max(left, right));
    }
};

算法4

时间复杂度：\(O(n)\)

代码

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int sum = nums[0], ans = nums[0];
        for (int i = 1; i < nums.size(); ++i) {
            if (sum < 0) sum = nums[i];
            else sum += nums[i];
            ans = max(ans, sum);
        }
        return ans;
    }
};