You are given a 0-indexed array nums consisting of positive integers.
You can do the following operation on the array any number of times:
Choose an integer i such that 0 <= i < nums.length - 1 and nums[i] <= nums[i + 1]. Replace the element nums[i + 1] with nums[i] + nums[i + 1] and delete the element nums[i] from the array.
Return the value of the largest element that you can possibly obtain in the final array.
Example 1:
Input: nums = [2,3,7,9,3]
Output: 21
Explanation: We can apply the following operations on the array:
- Choose i = 0. The resulting array will be nums = [5,7,9,3].
- Choose i = 1. The resulting array will be nums = [5,16,3].
- Choose i = 0. The resulting array will be nums = [21,3].
The largest element in the final array is 21. It can be shown that we cannot obtain a larger element.
Example 2:
Input: nums = [5,3,3]
Output: 11
Explanation: We can do the following operations on the array:
- Choose i = 1. The resulting array will be nums = [5,6].
- Choose i = 0. The resulting array will be nums = [11].
There is only one element in the final array, which is 11.
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 106
合并后数组中的最大元素。
给你一个下标从 0 开始、由正整数组成的数组 nums 。你可以在数组上执行下述操作 任意 次：
选中一个同时满足 0 <= i < nums.length - 1 和 nums[i] <= nums[i + 1] 的整数 i 。将元素 nums[i + 1] 替换为 nums[i] + nums[i + 1] ，并从数组中删除元素 nums[i] 。返回你可以从最终数组中获得的 最大 元素的值。
思路
思路是贪心。对于数组中的元素，只有满足 nums[i] <= nums[i + 1] 的时候，才能将 nums[i + 1] 替换为 nums[i] + nums[i + 1]，那么比较自然的想法是可以从数组的最右侧开始往左扫描，如果发现某个元素 nums[i] >= nums[i - 1]，就可以将 nums[i] 累加到 nums[i - 1] 上。这里我用一个变量 sum 来统计过程中能找到的最大的累加和。如果发现某个元素 nums[i] > sum，则说明从右往左扫的时候需要在这里停下，sum 要从 nums[i] 开始重新统计。
复杂度
时间O(n)
空间O(1)
代码
Java实现
class Solution {
public long maxArrayValue(int[] nums) {
int n = nums.length;
long sum = nums[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (sum >= nums[i]) {
sum += (long) nums[i];
} else {
sum = (long) nums[i];
}
}
return sum;
}
}