689. 三个无重叠子数组的最大和

给你一个整数数组 nums 和一个整数 k ，找出三个长度为 k 、互不重叠、且全部数字和（3 * k 项）最大的子数组，并返回这三个子数组。

以下标的数组形式返回结果，数组中的每一项分别指示每个子数组的起始位置（下标从 0 开始）。如果有多个结果，返回字典序最小的一个。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-sum-of-3-non-overlapping-subarrays
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution {

    private int[] getPreSum(int[] nums) {
        int[] preSum = new int[nums.length];
        preSum[0] = nums[0];
        for (int i = 1; i < nums.length; ++i) {
            preSum[i] = preSum[i - 1] + nums[i];
        }
        return preSum;
    }

    private int getRegionSum(int[] preSum, int left, int right) {
        return left == 0 ? preSum[right] : preSum[right] - preSum[left - 1];
    }

    public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
        if (nums == null || nums.length < 3 * k) {
            return new int[3];
        }
        int n = nums.length;
        int[] preSum = getPreSum(nums);
        int[] forward = new int[n];
        int[] behind = new int[n];
        forward[k - 1] = 0;
        for (int i = k; i < n; ++i) {
            forward[i] = getRegionSum(preSum, forward[i - 1], forward[i - 1] + k - 1) >=
                    getRegionSum(preSum, i - k + 1, i) ? forward[i - 1] : i - k + 1;
        }
        // 0 1 2 3
        behind[n - k] = n - k;
        for (int i = n - k - 1; i >= 0; --i) {
            behind[i] = getRegionSum(preSum, i, i + k - 1) >=
                    getRegionSum(preSum, behind[i + 1], behind[i + 1] + k - 1) ? i : behind[i + 1];
        }
        int[] ans = {};
        int max = Integer.MIN_VALUE;
        for (int i = k; i < n - 2 * k + 1; ++i) {
            int sum = getRegionSum(preSum, forward[i - 1], forward[i - 1] + k - 1) +
                    getRegionSum(preSum, i, i + k - 1) +
                    getRegionSum(preSum, behind[i + k], behind[i + k] + k - 1);
            if (sum > max) {
                max = sum;
                ans = new int[]{forward[i - 1], i, behind[i + k]};
            }
        }

        return ans;
    }
}

标签：重叠,nums,int,behind,689,数组,forward,preSum,getRegionSum
来源： https://www.cnblogs.com/tianyiya/p/15845664.html