[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]:
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]:
[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]:
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
[复杂度]:Time complexity: O() Space complexity: O()
[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] :
原文地址:https://www.cnblogs.com/immiao0319/p/9102541.html