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題目
Let’s call an array A a mountain if the following properties hold:
A.length >= 3
There exists some 0 < i < A.length - 1 such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]
Given an array that is definitely a mountain, return any i such that A[0] < A[1] < … A[i-1] < A[i] > A[i+1] > … > A[A.length - 1].
Example 1:
Input: [0,1,0]
Output: 1
Example 2:
Input: [0,2,1,0]
Output: 1
Note:
3 <= A.length <= 10000
0 <= A[i] <= 10^6
A is a mountain, as defined above.
分析
題意:給一個先遞增後遞減的數列,然後找出數列極值所在下標。
算法很簡單,無非就是遍歷一遍然後判斷當前值比前一個值大,比後一個值也大即可。
然而,實現是不夠在,更重要的是如何實現的又快又好。
解答
簡單實現(O(n))
class Solution {
public int peakIndexInMountainArray(int[] A) {
for(int i=1;i<A.length-1;i++){
if(A[i]>A[i-1]&&A[i]>A[i+1])
return i;
}
return -1;
}
}
去除冗餘條件(O(n))
class Solution {
public int peakIndexInMountainArray(int[] A) {
for (int i = 1; i + 1 < A.length; ++i){
if (A[i] > A[i + 1])
return i;
}
// for (int i = A.length - 1; i > 0; --i) if (A[i] > A[i - 1]) return i;
return 0;
}
}
由於我們是正向遍歷的,既然數組是升序,那麼在滿足A[i]>A[i+1]之前,必然滿足A[i]>A[i-1],因此可以省略A[i]>A[i-1](逆序遍歷則省略A[i]>A[i+1])
二分查找 (O(log n))
class Solution {
public int peakIndexInMountainArray(int[] A) {
int l = 0, r = A.length - 1, m;
while (l < r) {
m = (l + r) / 2;
if (A[m] < A[m + 1])
l = m + 1;
else
r = m;
}
return l;
}
}
更快的方法時黃金比例搜索
算法,有興趣的讀者可以查看最後一個解法