}returnright; } }; Github 同步地址: 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-t...
classSolution {publicintpeakIndexInMountainArray(int[] A) {if(A==null|| A.length==0){return0; }intleft = 0;intright = A.length-1;while(left<=right){intmid = (left+right)/2;if(A[mid-1]<A[mid] && A[mid]>A[mid+1]){returnmid; }elseif(A[mid-1]>A[mid]){ right= mid-...
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
"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
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}...
publicintpeakIndexInMountainArray(int[] A) {intindex=0;while(A[index] < A[index+1]) {index++; }returnindex; } 03 第三种解法 上面两种解法的时间复杂度都是O(N),我们还可以使用二分查找法,将时间复杂度降低为O(logN)。定义两个变量left、right,一个从0开始,一个从数组最后一位开始,每次取两者...
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; }
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]...