问题描述:
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).
解题思路:
这道题目参考的是zestypanda的解法
由于我们确定要找3个最大的不重复的子串,这样也就把整个字符串分成了3部分:左 右 中
可以这样分块:
左:[0, i]
右:[i, i+k-1]
中:[i+k, n-1]
我们可以先用动态规划算出左起和最大的长度为k的起始点的下标,存储在posLeft中
同样的右起和最大的下标放在posRight中。
注意初始值posLeft为0, 而PosRight 为n-k,因为一个从左遍历,一个从右遍历
填充好这两个数组后,就可以开始找最大的值了:
从第k个开始遍历,当当前和大于最大值时,将l, i,r放入返回数组中。
代码:
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; } };