LeetCode 050 Pow(x, n)
2016-05-30 09:24
316 查看
题目要求实现pow求幂函数。
使用
但要注意几个特殊情况。首先如果n是负数,那么可以通过
使用
pow(x,n)=pow(x,n/2)*pow(x,n/2)*pow(x,n%2)的性质求解。
但要注意几个特殊情况。首先如果n是负数,那么可以通过
pow(x,n)=1.0/pow(x,-n)求解。然后注意:当
n=INT_MIN时,
-n不存在!所以要特殊处理
n=INT_MIN,返回
pow(x,n/2)*pow(x,n/2)即可,先让n减半再取负数就行了。
相关文章推荐
- leetcode 179 Largest Number
- leetcode 24 Swap Nodes in Pairs
- leetcode 2 Add Two Numbers 方法1
- leetcode 2 Add Two Numbers 方法2
- leetcode----Longest Substring Without Repeating Characters
- [LeetCode]47 Permutations II
- [LeetCode]65 Valid Number
- [LeetCode]123 Best Time to Buy and Sell Stock III
- [LeetCode] String Reorder Distance Apart
- [LeetCode] Sliding Window Maximum
- [LeetCode] Find the k-th Smallest Element in the Union of Two Sorted Arrays
- [LeetCode] Determine If Two Rectangles Overlap
- [LeetCode] A Distance Maximizing Problem
- leetcode_linearList
- leetcode_linearList02
- 021-Merge Two Sorted Lists(合并两个排好序的单链表);leetcode
- LeetCode[Day 1] Two Sum 题解
- LeetCode[Day 2] Median of Two Sorted Arrays 题解
- LeetCode[Day 3] Longest Substring Without... 题解
- LeetCode [Day 4] Add Two Numbers 题解