We define an array is non-decreasing ifarray[i] <= array[i + 1]holds for everyi(1 <= i < n). Example 1: 4 1 Example 2: Input: [4,2,1] Output: False Explanation: You can't get a non-decreasing array by modify at most one element. Note: Thenbelongs to [1, 10,000]. 题...
01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第151题(顺位题号是665)。给定一个包含n个整数的数组,您的任务是通过修改最多1个元素来检查它是否可以变为非递减。如果array [i] <= array [i + 1],且0 <= i <n,则我们定义一个数组是非递减的。例如: 输入:[4,2,3] 输出:true 说明:可以修...
首先设置一个count变量,统计array[i] > array[i+1],如果count大于1就返回false; 对于不符合条件的元素,我们要给他重新赋值使整个数组可以满足单调递增 解法 Example1:Input:[4,2,3]Output:TrueExplanation:You could modify the first4to1togeta non-decreasing array. Example2:Input:[4,2,1]Output:FalseEx...
Given an arraynumswithnintegers, your task is to check if it could become non-decreasing by modifying at most one element. We define an array is non-decreasing ifnums[i] <= nums[i + 1]holds for everyi(0-based) such that (0 <= i <= n - 2). Example 1: Input: nums = [4,2...
Input:[4,2,1]Output:FalseExplanation:You can'tgeta non-decreasing array by modify at most one element. Note:Thenbelongs to [1, 10,000]. 自己琢磨的 :(简直无法忍受) 代码语言:javascript 复制 classSolution{publicbooleancheckPossibility(int[]nums){if(nums.length<=1){returntrue;}List<Integer>...
Explanation:You can’t get a non-decreasing array by modify at most one element. Note:Thenbelongs to [1, 10,000]. 思路 从index=1开始遍历数组,判断nums[i]是否小于nums[i-1],如果小,就进行下面的更换: i < 2 或者 nums[i] >= nums[i-2]:nums[i-1] = nums[i] ...
First, note that the order doesn't matter so we can sort the ai in non-decreasing order. Now, note that every interval's imbalance can be calculated with its largest and smallest value. We start adding the elements to sets from smallest to largest in order. Suppose we're adding the i...
(Pearson’sr = 0.8412,P < 0.0001) with IFNγ responses measured by cytokine bead array (CBA) of the same neoTCR in healthy donor cells generated at small-scale for initial product selection (Fig.2c). These data provide further validation of the neoTCR product selection process. ...
We define an array is non-decreasing ifarray[i] <= array[i + 1]holds for everyi(1 <= i < n). Example 1: 4 1 Example 2: Input: [4,2,1] Output: False Explanation: You can't get a non-decreasing array by modify at most one element. ...
https://leetcode.com/problems/non-decreasing-array/discuss/106826/JavaC%2B%2B-Simple-greedy-like-solution-with-explanation 用count来表示non decreasing的subarray的数量,如果是1直接return true,如果大于2直接返回false。 如果是2,说明是两段。分两种情况,第一种要求转折点index有index + 1 >= index - 1...