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}
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; } }return-1; } };
思路 A[i]为数组A的最大值,然后返回其在A中的索引即可 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
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。
"852. Peak Index in a Mountain Array Easy" 方法一:二分查找 官方答案: java class Solution { public int peakIndexInMountainArray(int[] A) { int lo = 0, h
3 <= A.length <= 10000 0 <= A[i] <= 10^6 A is a mountain, as defined above.二分法:class Solution{ public: int peakIndexInMountainArray(vector<int>& a){ int beg = 1,end = a.size(); int mid = (beg + end) / 2; while(beg <= end){ if(a[mid] < a[mid - 1]){ ...
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; }
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 ...
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]...