辣雞劉的Leetcode之旅8【Add Binary，Sqrt(x)，Merge Sorted Array】

67. Add Binary
    題目描述;

Given two binary strings, return their sum (also a binary string).
The input strings are both non-empty and contains only characters 1 or 0.
Example 1:
Input: a = "11", b = "1"
Output: "100"
Example 2:
Input: a = "1010", b = "1011"
Output: "10101"

簡單描述：二進制相加
代碼如下：

class Solution:
    def addBinary(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: str
        """
        print(bin(int(a,2)+int(b,2))[2:])
        return bin(int(a,2)+int(b,2))[2:]
    
s=Solution()
s.addBinary("1010","1011")

python的函數真他孃的多。
本題中用到的：

bin() 返回一個整數 int 或者長整數 long int 的二進制表示。
示例如下：
>>>bin(10)
'0b1010'
>>> bin(20)
'0b10100'

上面的例子也說明了代碼中爲什麼用到了[2:]，就是從第3位開始；
然後讓我們重新認識以下int這個函數：

int(x, base=10)
x -- 字符串或數字。
base -- 進制數，默認十進制。
舉例如下：
>>>int()               # 不傳入參數時，得到結果0
0
>>> int(3)
3
>>> int(3.6)
3
>>> int('12',16)        # 如果是帶參數base的話，12要以字符串的形式進行輸入，12 爲 16進制
18
>>> int('0xa',16)  
10  
>>> int('10',8)  
8

69. Sqrt(x)

題目描述：

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.
Example 1:
Input: 4
Output: 2
Example 2:
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

翻譯：求所給數字的平方根；
我做的第一種方法：

class Solution:
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
		mid=x//2
        if x == 0 or x == 1:
            return x
        for i in range(mid+1):
            if i * i <= x and (i + 1) * (i + 1) > x:
                print(i)
                return i

這種方法在編譯器上可以AC，但是到了LeetCode的平臺就提示：

351 / 1017 test cases passed.
Status: Time Limit Exceeded

也就是說時間複雜度太大了，畢竟是一個一個遍歷。
第二種方法【二分思想】：

class Solution:
    def mySqrt(self, x):
        """
        :type x: int
        :rtype: int
        """
        low,high=0,x
        while low <= high:
            mid = (low + high) / 2
            if mid ** 2 <= x:
                low = mid + 1
            else:
                high = mid - 1
        return low - 1


s=Solution()
s.mySqrt(4)

88. Merge Sorted Array

題目描述：
Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:

The number of elements initialized in nums1 and nums2 are m and n respectively.
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2.
就是把兩個已經排好序的數組，在不借助其他內存的基礎上合併到一起（也是排好序的）
Example:

Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3
Output: [1,2,2,3,5,6]

我的解法：
起初我是這樣寫的：

class Solution:
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: void Do not return anything, modify nums1 in-place instead.
        """
        for i in range(len(nums2)):
            nums1.append(nums2[i])
        nums1.sort()

s=Solution()
s.merge([1,2,3,0,0,0],3,[2,5,6],3)

可是輸出有0沒法處理；
最後這樣寫的：

for i in range(len(nums2)):
     nums1[i+m]=nums2[i]
nums1.sort()

還有一種寫法，值得學習，利用了堆棧的思維；

while(len(nums1)>m):
     nums1.pop(m)
while(len(nums2)>n):
     nums2.pop(n)
for i in nums2:
     nums1.append(i)
nums1.sort()