leetcode【845】Longest Mountain in Array 写在最前面:不得不说,python对于数据结构的处理非常自然,几乎就是你怎么想的直接写出来就可以,不用太过在意转化,代价就是,性能较差 leetcode【845】Longest Mounta in Array Let's call any (contiguous) subarray B (of A) a mountain if the following properties...
There exists some0 < i < B.length - 1such thatB[0] < B[1] < ... B[i-1] < B[i] > B[i+1] > ... > B[B.length - 1] (Note that B could be any subarray of A, including the entire array A.) Given an arrayAof integers, return the length of the longestmountain. Ret...
res= max(res, right - left +1); } }returnres; } }; Github 同步地址: https://github.com/grandyang/leetcode/issues/845 参考资料: https://leetcode.com/problems/longest-mountain-in-array/ https://leetcode.com/problems/longest-mountain-in-array/discuss/176952/Java-1-pass-and-O(1)-space...
int longestMountain(int* A, int ASize){ if (ASize < 3) { return 0; } bool decrease = false; int i, max = 0, temp = 0; for (i = 0; i < ASize-1; i++) { // 若处于递增状态 if (A[i + 1] > A[i]) { // 结束情况2:前面数字处于递减状态,而当前处于递增状态 if (...
大致思路是把每一个位置当成起点,然后先找up上升趋势的值,然后找下降趋势的值,这样就能找到一个mountain array,我们更新res来找到所有mountain array中最长的array的长度。 代码 class Solution { public: int longestMountain(vector<int>& A) { int n=A.size(); ...
数列中找最长的“山形”subarray,即先递增再递减的子列。要求1-pass,O(1)的空间。 难度:medium 算法:DP 思路: 这题本来可以用两个指针按顺序找,因为最大山形数列不会有重叠。 由于刚做过另一道题LeetCode 53 求和最大的子列,也就同样用了动态规划的思路去做。这题的DP稍微复杂一点,除了记录山形数列还要记录...
给出一个整数数组arr,返回最长山脉子数组的长度。如果不存在山脉子数组,返回0。 示例1: 输入:arr = [2,1,4,7,3,2,5]输出:5解释:最长的山脉子数组是 [1,4,7,3,2],长度为 5。 示例2: 输入:arr = [2,2,2]输出:0解释:不存在山脉子数组。
Can you solve this real interview question? Longest Mountain in Array - You may recall that an array arr is a mountain array if and only if: * arr.length >= 3 * There exists some index i (0-indexed) with 0 < i < arr.length - 1 such that: * arr[0] <
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