<gh_stars>10-100
# Implement pow(x, n), which calculates x raised to the power n (xn).

# Example 1:
# Input: 2.00000, 10
# Output: 1024.00000

# Example 2:
# Input: 2.10000, 3
# Output: 9.26100

# Example 3:
# Input: 2.00000, -2
# Output: 0.25000
# Explanation: 2-2 = 1/22 = 1/4 = 0.25

# Note:
# -100.0 < x < 100.0
# n is a 32-bit signed integer, within the range [−2^31, 2^31 − 1]

class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        # M0. 自带库
        # return x ** n

        # M1. 蛮力算法 O(n)
        # if x == 0:
        #     return 0
        # if n < 0:
        #     x = 1/x
        #     n = -n

        # res = 1
        # # Memory Exceeded for range
        # # TimeLimitExceeded for xrange
        # for i in range(n):
        # # for i in xrange(n):
        #     res *= x
        # return res

        # M2. 二分法 O(lgn)
        if x == 0:
            return 0
        if n < 0:
            x = 1/x
            n = -n

        res = 1
        while n >= 1:
            if n%2 == 1:
                res *= x
            x *= x
            n >>= 1
        return res
# Implement pow(x, n), which calculates x raised to the power n (xn).

# Example 1:
# Input: 2.00000, 10
# Output: 1024.00000

# Example 2:
# Input: 2.10000, 3
# Output: 9.26100

# Example 3:
# Input: 2.00000, -2
# Output: 0.25000
# Explanation: 2-2 = 1/22 = 1/4 = 0.25

# Note:
# -100.0 < x < 100.0
# n is a 32-bit signed integer, within the range [−2^31, 2^31 − 1]

class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        # M0. 自带库
        # return x ** n

        # M1. 蛮力算法 O(n)
        # if x == 0:
        #     return 0
        # if n < 0:
        #     x = 1/x
        #     n = -n

        # res = 1
        # # Memory Exceeded for range
        # # TimeLimitExceeded for xrange
        # for i in range(n):
        # # for i in xrange(n):
        #     res *= x
        # return res

        # M2. 二分法 O(lgn)
        if x == 0:
            return 0
        if n < 0:
            x = 1/x
            n = -n

        res = 1
        while n >= 1:
            if n%2 == 1:
                res *= x
            x *= x
            n >>= 1
        return res
# Implement pow(x, n), which calculates x raised to the power n (xn). # Example 1: # Input: 2.00000, 10 # Output: 1024.00000 # Example 2: # Input: 2.10000, 3 # Output: 9.26100 # Example 3: # Input: 2.00000, -2 # Output: 0.25000 # Explanation: 2-2 = 1/22 = 1/4 = 0.25 # Note: # -100.0 < x < 100.0 # n is a 32-bit signed integer, within the range [−2^31, 2^31 − 1] :type x: float         :type n: int         :rtype: float # M0. 自带库 # return x ** n # M1. 蛮力算法 O(n) # if x == 0: #     return 0 # if n < 0: #     x = 1/x #     n = -n # res = 1 # # Memory Exceeded for range # # TimeLimitExceeded for xrange # for i in range(n): # # for i in xrange(n): #     res *= x # return res # M2. 二分法 O(lgn)