Max Sum of Subarray with size k, 相当于Easy难度，我说用一个sum array存sum，然后做减法就行。中国小哥说让优化空间，于是说可以只用两个数。
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.
The question asks for three non-overlapping intervals with maximum sum of all 3 intervals. If the middle interval is [i, i+k-1], where k <= i <= n-2k, the left interval has to be in subrange [0, i-1], and the right interval is from subrange [i+k, n-1].
So the following solution is based on DP.
posLeft[i] is the starting index for the left interval in range [0, i];
posRight[i] is the starting index for the right interval in range [i, n-1];
Then we test every possible starting index of middle interval, i.e. k <= i <= n-2k, and we can get the corresponding left and right max sum intervals easily from DP. And the run time is O(n).
Caution. In order to get lexicographical smallest order, when there are two intervals with equal max sum, always select the left most one. So in the code, the if condition is ">= tot" for right interval due to backward searching, and "> tot" for left interval. Thanks to @lee215 for pointing this out!
class Solution {
public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
int n = nums.length, maxsum = 0;
int[] sum = new int[n+1], posLeft = new int[n], posRight = new int[n], ans = new int[3];
for (int i = 0; i < n; i++) sum[i+1] = sum[i]+nums[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
posRight[n-k] = n-k;
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[0] = l; ans[1] = i; ans[2] = r;
}
}
return ans;
}
}