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 All in One 题目讲解汇总(持续更新中...)...
class Solution { public int peakIndexInMountainArray(int[] A) { int lo = 0, hi = A.length - 1; while (lo < hi) { int mi = lo + (hi - lo) / 2; if (A[mi] < A[mi + 1]) lo = mi + 1; else hi = mi; } return lo; } } Time Complexity: O(logN) Space Complexity...
def peakIndexInMountainArray(self, A): """ :type A: List[int] :rtype: int """ return A.index(max(A)) 1. 2. 3. 4. 5. 6. 7.
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;
class Solution{ public: int peakIndexInMountainArray(vector<int>& a){ int max_elem = *max_element(a.begin(), a.end()); int pos; for(pos = 0; pos < a.size(); ++pos){ if(a[pos] == max_elem) break; } return pos; } };...
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]...
原题链接在这里:https://leetcode.com/problems/peak-index-in-a-mountain-array/ 题目: Let's call an arrayAamountainif 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[...
今天介绍的是LeetCode算法题中Easy级别的第199题(顺位题号是852)。如果以下属性成立,我们将数组A称为山: A.length> = 3。 存在一个i(0 < i < A.length-1),使得A[0] <A[1] <... A[i-1] < A[i] > A[i + 1]> ...> A[A.length - 1]。