689.三个无重叠子数组的最大和

题目描述

三个无重叠子数组的最大和

思路

滑动窗口

使用3个大小为k的滑动窗口。设$su{m}_1$为第一个滑动窗口的元素和，该滑动窗口从[0, k-1]开始；$su{m}_2$为第二个滑动窗口的元素和，该滑动窗口从[k, 2k-1]开始；$su{m}_3$位第三个滑动窗口的元素和，该滑动窗口从[2k, 3*k-1]开始。
同时向右滑动三个滑动窗口，每次滑动时，计算当前$maxSu{m}_{12}$与$su{m}_3$之和。统计这一过程中的$maxSu{m}_{12}+su{m}_3$的最大值及对应位置。
而$maxSu{m}_{12}$可以由$maxSu{m}_1$和$su{m}_2$的和来求。
题目还要求使用最小字典序，因为是从左往右遍历，所以当且仅当元素和大于最大元素和时，才修改最大元素和以及对应位置。

Java实现

class Solution {
    
    
    public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
    
    
        int[] ans = new int[3];
        int sum1 = 0, maxSum1 = 0, maxSum1Idx = 0;
        int sum2 = 0, maxSum12 = 0, maxSum12Idx1 = 0, maxSum12Idx2 = 0;
        int sum3 = 0, maxTotal = 0;
        for (int i = k * 2; i < nums.length; ++i) {
    
    
            sum1 += nums[i - k * 2];
            sum2 += nums[i - k];
            sum3 += nums[i];
            if (i >= k * 3 - 1) {
    
    
                if (sum1 > maxSum1) {
    
    
                    maxSum1 = sum1;
                    maxSum1Idx = i - k * 3 + 1;
                }
                if (maxSum1 + sum2 > maxSum12) {
    
    
                    maxSum12 = maxSum1 + sum2;
                    maxSum12Idx1 = maxSum1Idx;
                    maxSum12Idx2 = i - k * 2 + 1;
                }
                if (maxSum12 + sum3 > maxTotal) {
    
    
                    maxTotal = maxSum12 + sum3;
                    ans[0] = maxSum12Idx1;
                    ans[1] = maxSum12Idx2;
                    ans[2] = i - k + 1;
                }
                sum1 -= nums[i - k * 3 + 1];
                sum2 -= nums[i - k * 2 + 1];
                sum3 -= nums[i - k + 1];
            }
        }
        return ans;
    }
}

Python实现

class Solution:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
        ans = []
        sum1, maxSum1, maxSum1Idx = 0, 0, 0
        sum2, maxSum12, maxSum12Idx = 0, 0, ()
        sum3, maxTotal = 0, 0
        for i in range(k * 2, len(nums)):
            sum1 += nums[i - k * 2]
            sum2 += nums[i - k]
            sum3 += nums[i]
            if i >= k * 3 - 1:
                if sum1 > maxSum1:
                    maxSum1 = sum1
                    maxSum1Idx = i - k * 3 + 1
                if maxSum1 + sum2 > maxSum12:
                    maxSum12 = maxSum1 + sum2
                    maxSum12Idx = (maxSum1Idx, i - k * 2 + 1)
                if maxSum12 + sum3 > maxTotal:
                    maxTotal = maxSum12 + sum3
                    ans = [*maxSum12Idx, i - k + 1]
                sum1 -= nums[i - k * 3 + 1]
                sum2 -= nums[i - k * 2 + 1]
                sum3 -= nums[i - k + 1]
        return ans