LeetCode 50 Pow(x, n)
2018-01-24 22:44
337 查看
递归分解成子问题解决:
class Solution { public double myPow(double x, int n) { if(n==0) return 1; if(n>0){ if(n%2==0) return myPow(x*x,n/2); else return myPow(x*x,n/2)*x; } if(n<0) { //必须-(n/2) 不可以-n/2会溢出 //如果n是-2147483648 //-n就是2147483648 //溢出 if(n%2==0) return 1/myPow(x*x,-(n/2)); else return 1/myPow(x*x,-(n/2))/x; } return 0; } }
相关文章推荐
- 【Leetcode】算法题50 Pow(x,n)
- Leetcode-50: Pow(x,n)
- 【LeetCode】50、pow(x,n)
- leetcode-50-pow
- LeetCode 50 - Pow(x, n)
- LeetCode50 pow(x, n)
- Leetcode 50 Pow(x,n) (求x的n次方)
- Leetcode50 Pow(x,n)
- Leetcode50 Pow(x, n)
- Leetcode 50 Pow(x, n) 两种方式求解
- leetcode[50]Pow(x, n)
- 【LeetCode】LeetCode50 Pow(x, n)
- [leetcode 50]pow(x,n)
- LeetCode_50_Pow(x, n)
- leetcode 50: Pow(x, n)
- LeetCode-50 Pow(x, n)
- LeetCode 50 — Pow(x, n)(C++ Java Python)
- LeetCode 50 Pow(x, n)(Math、Binary Search)(*)
- LeetCode 50 Pow(x, n)
- leetcode_question_50 Pow(x, n)