19.2.8 [LeetCode 53] Maximum Subarray

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Example:

Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

Follow up:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

题意

求连续子集的最大和

题解

其实我觉得这题挺难的|||

首先是O(n)解法

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int e = 1, n = nums.size(), ans = nums[0], now = nums[0];
        while (e < n) {
            now = max(now + nums[e], nums[e]);
            ans = max(now, ans);
            e++;
        }
        return ans;
    }
};

然后是分治算法

class Solution {
public:
    int maxSubRange(vector<int>&nums, int x, int y) {
        if (x == y)return nums[x];
        int mid = (x + y) / 2;
        int lmax = maxSubRange(nums, x, mid), rmax = maxSubRange(nums, mid + 1, y);
        int tmp = 0, mmax = 0;
        for (int i = mid - 1; i >= x; i--) {
            tmp += nums[i];
            mmax = max(tmp, mmax);
        }
        tmp = mmax;
        for (int i = mid; i <= y; i++) {
            tmp += nums[i];
            mmax = max(tmp, mmax);
        }
        return max(mmax, max(lmax, rmax));
    }
    int maxSubArray(vector<int>& nums) {
        int e = 1, n = nums.size(), ans = nums[0], now = nums[0];
        while (e < n) {
            now = max(now + nums[e], nums[e]);
            ans = max(now, ans);
            e++;
        }
        return ans;
    }
};