[LeetCode] 689. Maximum Sum of 3 Non-Overlapping Subarrays 三个非重叠子数组的最大和

In a given array nums of positive integers, find three non-overlapping subarrays with maximum sum.

Each subarray will be of size k, and we want to maximize the sum of all 3*k entries.

Return the result as a list of indices representing the starting position of each interval (0-indexed). If there are multiple answers, return the lexicographically smallest one.

Example:

Input: [1,2,1,2,6,7,5,1], 2
Output: [0, 3, 5]
Explanation: Subarrays [1, 2], [2, 6], [7, 5] correspond to the starting indices [0, 3, 5].
We could have also taken [2, 1], but an answer of [1, 3, 5] would be lexicographically larger. 

Note:

• nums.length will be between 1 and 20000.
• nums[i] will be between 1 and 65535.
• k will be between 1 and floor(nums.length / 3).

给一个由正数组成的数组，找三个长度为k的不重叠的子数组，使得三个子数组的数字之和最大。

解法： DP，思路类似于123. Best Time to Buy and Sell Stock III，先分别从左和右两个方向求出每一个位置i之前的长度为k的元素和最大值，这样做的好处是之后想要得到某一位置的最大和时能马上知道。然后在用一个循环找出三段的最大和。

Java:

class Solution {
    public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
        int n = nums.length, maxsum = 0;
        int[] sum = new int[n+1], posLeft = new int[n], posRight = new int[n], ans = new int[3];
        for (int i = 0; i < n; i++) sum[i+1] = sum[i]+nums[i];
        // DP for starting index of the left max sum interval
        for (int i = k, tot = sum[k]-sum[0]; i < n; i++) {
            if (sum[i+1]-sum[i+1-k] > tot) {
                posLeft[i] = i+1-k;
                tot = sum[i+1]-sum[i+1-k];
            }
            else
                posLeft[i] = posLeft[i-1];
        }
        // DP for starting index of the right max sum interval
       // caution: the condition is ">= tot" for right interval, and "> tot" for left interval
        posRight[n-k] = n-k;
        for (int i = n-k-1, tot = sum[n]-sum[n-k]; i >= 0; i--) {
            if (sum[i+k]-sum[i] >= tot) {
                posRight[i] = i;
                tot = sum[i+k]-sum[i];
            }
            else
                posRight[i] = posRight[i+1];
        }
        // test all possible middle interval
        for (int i = k; i <= n-2*k; i++) {
            int l = posLeft[i-1], r = posRight[i+k];
            int tot = (sum[i+k]-sum[i]) + (sum[l+k]-sum[l]) + (sum[r+k]-sum[r]);
            if (tot > maxsum) {
                maxsum = tot;
                ans[0] = l; ans[1] = i; ans[2] = r;
            }
        }
        return ans;
    }
}

Python:　　

class Solution(object):
    def maxSumOfThreeSubarrays(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        n = len(nums)
        accu = [0]
        for num in nums:
            accu.append(accu[-1]+num)

        left_pos = [0] * n
        total = accu[k]-accu[0]
        for i in xrange(k, n):
            if accu[i+1]-accu[i+1-k] > total:
                left_pos[i] = i+1-k
                total = accu[i+1]-accu[i+1-k]
            else:
                left_pos[i] = left_pos[i-1]

        right_pos = [n-k] * n
        total = accu[n]-accu[n-k]
        for i in reversed(xrange(n-k)):
            if accu[i+k]-accu[i] > total:
                right_pos[i] = i;
                total = accu[i+k]-accu[i]
            else:
                right_pos[i] = right_pos[i+1]

        result, max_sum = [], 0
        for i in xrange(k, n-2*k+1):
            left, right = left_pos[i-1], right_pos[i+k]
            total = (accu[i+k]-accu[i]) +                     (accu[left+k]-accu[left]) +                     (accu[right+k]-accu[right])
            if total > max_sum:
                max_sum = total
                result = [left, i, right]
        return result

C++:

class Solution {
public:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int n = nums.size(), maxsum = 0;
        vector<int> sum = {0}, posLeft(n, 0), posRight(n, n-k), ans(3, 0);
        for (int i:nums) sum.push_back(sum.back()+i);
       // DP for starting index of the left max sum interval
        for (int i = k, tot = sum[k]-sum[0]; i < n; i++) {
            if (sum[i+1]-sum[i+1-k] > tot) {
                posLeft[i] = i+1-k;
                tot = sum[i+1]-sum[i+1-k];
            }
            else
                posLeft[i] = posLeft[i-1];
        }
        // DP for starting index of the right max sum interval
        // caution: the condition is ">= tot" for right interval, and "> tot" for left interval
        for (int i = n-k-1, tot = sum[n]-sum[n-k]; i >= 0; i--) {
            if (sum[i+k]-sum[i] >= tot) {
                posRight[i] = i;
                tot = sum[i+k]-sum[i];
            }
            else
                posRight[i] = posRight[i+1];
        }
        // test all possible middle interval
        for (int i = k; i <= n-2*k; i++) {
            int l = posLeft[i-1], r = posRight[i+k];
            int tot = (sum[i+k]-sum[i]) + (sum[l+k]-sum[l]) + (sum[r+k]-sum[r]);
            if (tot > maxsum) {
                maxsum = tot;
                ans = {l, i, r};
            }
        }
        return ans;
    }
};

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[LeetCode] 123. Best Time to Buy and Sell Stock III 买卖股票的最佳时间 III

原文地址：https://www.cnblogs.com/lightwindy/p/9692653.html