三個無重疊子陣列的最大和
阿新 • • 發佈:2021-10-29
給定陣列nums由正整陣列成,找到三個互不重疊的子陣列的最大和。
每個子陣列的長度為k,我們要使這3*k個項的和最大化。
返回每個區間起始索引的列表(索引從 0 開始)。如果有多個結果,返回字典序最小的一個。
來源:力扣(LeetCode)
連結:https://leetcode-cn.com/problems/maximum-sum-of-3-non-overlapping-subarrays
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心之所向,素履以往 生如逆旅,一葦以航import java.util.Arrays; class Solution { private static int getSum(int[] preSum, int l, int r) { if (l == 0) { return preSum[r]; } return preSum[r] - preSum[l - 1]; } public static int[] maxSumOfThreeSubarrays(int[] nums, int k) { if (nums == null || nums.length == 0) { return new int[3]; } int n = nums.length; int[] preSum = new int[n]; preSum[0] = nums[0]; for (int i = 1; i < n; ++i) { preSum[i] = preSum[i - 1] + nums[i]; } /** * forward[i]:直接到i左側最優的起始位置 */ int[] forward = new int[n]; /** * behind[i]:截止到i右側最優的起始位置 */ int[] behind = new int[n]; // 0 1 2 3 4 5 forward[k - 1] = 0; for (int i = k; i < n; ++i) { forward[i] = getSum(preSum, i - k + 1, i) > getSum(preSum, forward[i - 1], forward[i - 1] + k - 1) ? i - k + 1 : forward[i - 1]; } behind[n - k] = n - k; for (int i = n - k - 1; i >= 0; --i) { behind[i] = getSum(preSum, i, i + k - 1) >= getSum(preSum, behind[i + 1], behind[i + 1] + k - 1) ? i : behind[i + 1]; } int[] ret = new int[3]; int max = 0; for (int i = k; i <= n - 2 * k; ++i) { int sum = getSum(preSum, forward[i - 1], forward[i - 1] + k - 1) + getSum(preSum, behind[i + k], behind[i + k] + k - 1) + getSum(preSum, i, i + k - 1); if (sum > max) { max = sum; ret = new int[]{forward[i - 1], i, behind[i + k]}; } } return ret; } public static void main(String[] args) { int[] arr = new int[]{1, 2, 1, 2, 6, 7, 5, 1}; int k = 2; int[] ints = maxSumOfThreeSubarrays(arr, k); Arrays.stream(ints).forEach(System.out::println); } }