689. Maximum Sum of 3 Non-Overlapping Subarrays三个不重合数组的求和最大值

［抄题］：

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.

[暴力解法]：

时间分析：

空间分析：

[优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

note中已经提示了length，就只需要考虑k k&length的关系就了

把“前i项”初始化为“第i项”，方便直接做差

for (int i = 1; i <= n; i++) {

            sums[i] = sums[i - 1] + nums[i - 1];

        }

［思维问题］：

不知道为什么要用DP:每次都保存之前一组的状态，然后一个个向前更新和比价。

求一组固定为k长度的数组时可用。

//总和=本组和+之前组的和=本组最后之和-本组第一之和+之前的（从j - k开始的）dp求和值

int curSum = sums[j] - sums[j - k] + dp[i - 1][j - k];

［英文数据结构或算法，为什么不用别的数据结构或算法］：

dp数组里存储了结果，可以通过不断输入index来把结果取出来：

int index = n;

        for (int i = 2; i >= 0; i--) {

            res[i] = pos[i + 1][index];

            System.out.println("index = " +index);

            System.out.println("res[i] = pos[i + 1][index] = " +res[i]);

            index = res[i];

            System.out.println("index = " +index);

            System.out.println("----------------");

        }

［一句话思路］：

按照第123组来操作，

 总和=本组和+之前所有组的和

［输入量］：空：正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

序列型dp所有的有关数组、有关二维数组都要增加1个单位，调用的时候也要+1,因为第一位拿来初始化了。不初始化就是默认为0

［二刷］：

发现把第0位给去掉了不知道为何：

[1,2,1,2,6,7,5,1]

2

i = 1

i - 1 = 0

nums[i - 1] = 1

sum[i - 1] = 0

sum[i - 1] = 1

---------------

i = 2

i - 1 = 1

nums[i - 1] = 2

sum[i - 1] = 0

sum[i - 1] = 2

---------------

i = 3

i - 1 = 2

nums[i - 1] = 1

sum[i - 1] = 0

sum[i - 1] = 1

---------------

i = 4

i - 1 = 3

nums[i - 1] = 2

sum[i - 1] = 0

sum[i - 1] = 2

---------------

i = 5

i - 1 = 4

nums[i - 1] = 6

sum[i - 1] = 0

sum[i - 1] = 6

---------------

i = 6

i - 1 = 5

nums[i - 1] = 7

sum[i - 1] = 0

sum[i - 1] = 7

---------------

i = 7

i - 1 = 6

nums[i - 1] = 5

sum[i - 1] = 0

sum[i - 1] = 5

---------------

i = 8

i - 1 = 7

nums[i - 1] = 1

sum[i - 1] = 0

sum[i - 1] = 1

---------------

［三刷］：

［四刷］：

［五刷］：

[五分钟肉眼debug的结果]：

［总结］：

dp是存储一组状态的，可以拿来调用

［复杂度］：Time complexity: O(n) Space complexity: O(n)

［算法思想：迭代/递归/分治/贪心］：

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

[代码风格] ：

[是否头一次写此类driver funcion的代码] ：

class Solution {

    public int[] maxSumOfThreeSubarrays(int[] nums, int k) {

        //ini: res[3], pos[4][n + 1], dp[4][n + 1]

        int n = nums.length;

        int[] res = new int[3];

        int[] sum = new int[n + 1];

        int[][] pos = new int[4][n + 1];

        int[][] dp = new int[4][n + 1];

        //cc

        if (nums == null || nums.length < 3 * k) return res;

        //ini:sum

        for (int i = 1; i <= n; i++) {

            int j = i - 1;

            System.out.println("i = "+i);

            System.out.println("i - 1 = "+j);

            System.out.println("nums[i - 1] = "+nums[i - 1]);

            System.out.println("sum[i - 1] = "+sum[i - 1]);

            sum[i - 1] = sum[i - 1] + nums[i - 1];

            System.out.println("sum[i - 1] = "+sum[i - 1]);

            System.out.println("---------------");

        }

        for (int i = 1; i <= 3; i++) {

            for (int j = k * i; j <= n; j++) {

                int curSum = sum[j] - sum[j - k] + dp[i - 1][j - k];

                if (curSum > dp[i][j - 1]) {

                    dp[i][j] = curSum;

                    pos[i][j] = j - k;

                }else {

                    dp[i][j] = dp[i][j - 1];

                    pos[i][j] = pos[i][j - 1];

                }

            }

        }

        //retrieve the answer

        int index = n;

        for (int i = 2; i >= 0; i--) {

            //

            res[i] = pos[i + 1][index];

            index = res[i];

        }

        //return

        return res;

    }

}

巴特西

689. Maximum Sum of 3 Non-Overlapping Subarrays三个不重合数组的求和最大值

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