publicintpeakIndexInMountainArray(int[] A){intn = A.length;for(inti=0; i<n-1; i++) {if(A[i] < A[i+1]) {continue; }else{returni; } }return0; } 03 第二种解法 思路和上面一样,也是直接遍历数组元素,进行比较,找到山顶。 publicintpeakIndexInMountainArray(int[] A) {intindex=0;...
Else, peak must be within mid left side including mid. r = mid. Time Complexity: O(logn). n = A.length. Space: O(1). AC Java: 1classSolution {2publicintpeakIndexInMountainArray(int[] A) {3if(A ==null|| A.length == 0){4return-1;5}67intl = 0;8intr = A.length - 1;9...
代码实现 class Solution: def peakIndexInMountainArray(self, A): """ :type A: List[int] :rtype: int """ return A.index(max(A)) 1. 2. 3. 4. 5. 6. 7.
LeetCode 852. Peak Index in a Mountain Array 2019-12-21 09:43 −原题链接在这里:https://leetcode.com/problems/peak-index-in-a-mountain-array/ 题目: Let's call an array A a mountain if the following proper... Dylan_Java_NYC
LeetCode题解之Peak Index in a MountainArray 1 题目描述 2、问题分析 直接从后向前遍历,找到 A[i] > A[i-1] 即可。 3.代码 1intpeakIndexInMountainArray(vector<int>&A) {2inti = A.size() -1;3while( i >0){4if( A[i] > A[i-1]){5returni;6}7i--;8}9}...
3 <= A.length <= 10000 0 <= A[i] <= 10^6 A是如上定义的山脉 解法: classSolution{public:intpeakIndexInMountainArray(vector<int>& A){intsz = A.size();for(inti =1; i < sz; i++){if(A[i] < A[i-1]){returni-1;
https://github.com/grandyang/leetcode/issues/852 类似题目: Find Peak Element 参考资料: https://leetcode.com/problems/peak-index-in-a-mountain-array/ https://leetcode.com/problems/peak-index-in-a-mountain-array/discuss/139848/C%2B%2BJavaPython-Better-than-Binary-Search ...
classSolution {public:intpeakIndexInMountainArray(vector<int>&A) {intlen=A.size();for(inti=1;i<len-1;++i){if(A[i]>A[i-1] && A[i]>A[i+1]){returni; } }if(len>1&& A[len-1]>A[len-2])returnlen-1;return0; }
LeetCode 852 Peak Index in a Mountain Array 解题报告 题目要求 Let's call an arrayAa mountain if the following properties hold: A.length >= 3 There exists some0 < i < A.length - 1such thatA[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]...
1.Leetcode_easy_852. Peak Index in a Mountain Array; 2.grandyang; 3.discuss; 完 各美其美,美美与共,不和他人作比较,不对他人有期待,不批判他人,不钻牛角尖。 心正意诚,做自己该做的事情,做自己喜欢做的事情,安静做一枚有思想的技术媛。