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］：

1. 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］：

1. 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;
    }
}