2021.08.20-力扣刷题（70，69、83、88、100）

本文主要是介绍2021.08.20-力扣刷题（70，69、83、88、100），对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

70. 爬楼梯

链接：https://leetcode-cn.com/problems/climbing-stairs/

方法一：动态规划

class Solution:
    def climbStairs(self, n: int) -> int:
        dp = [0 for _ in range(n+1)]
        
        if n<=2: return n

        for i in range(3, n+1):
            dp[1] = 1
            dp[2] = 2
            dp[i] = dp[i-1] + dp[i-2]
        
        return dp[-1]

  方法二：状态压缩

class Solution:
    def climbStairs(self, n: int) -> int:
        if n <= 2:
            return n
        
        dp = [1, 2]
        for i in range(3, n+1):
            ans = dp[0] + dp[1]
            dp[0], dp[1] = dp[1], ans
        
        return dp[1]

 69. x 的平方根

链接：https://leetcode-cn.com/problems/sqrtx/

class Solution:
    def mySqrt(self, x: int) -> int:
        if x < 2: return x
        left = 1
        right = x//2

        while left <= right:
            mid = left + (right - left) // 2

            if mid ** 2 > x:
                right = mid -1
            
            else:
                left = mid + 1

        return right
            

 83. 删除排序链表中的重复元素

链接：https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list/

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        start = head
        while start:
            cur = start.next
            while cur:
                if cur.val == start.val:
                    cur = cur.next
                else:
                    break
            start.next = cur
            start = start.next
        return head
            

88. 合并两个有序数组

链接：https://leetcode-cn.com/problems/merge-sorted-array/

方法一：双指针（但是通不过，在本地编辑器里能通过）

class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        """
        Do not return anything, modify nums1 in-place instead.
        """

        ans = []
        if m == 0: ans = nums2
        if n == 0: ans = nums1[:m]
        i=0
        j=0
        k=0
        while i<m and j<n:
            if nums1[i] <= nums2[j]:
                ans.append(nums1[i])
                i += 1
            else:
                ans.append(nums2[j])
                j += 1
            k += 1

        if i < m:
            ans.extend(nums1[i:])
        elif j < n:
            ans.extend(nums2[j:])

        nums1[:] = sorted(ans)

方法二：切片后排序

最直观的方法是先将数组nums2放进数组nums1[m:] 的尾部，然后直接对整个数组进行排序。

class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        """
        Do not return anything, modify nums1 in-place instead.
        """

        nums1[m:]= nums2
        nums1.sort()
        

100. 相同的树

 链接：https://leetcode-cn.com/problems/same-tree/submissions/

方法：深度优先dfs

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
        if not p and not q: return True
        if p == None or q == None: return False
        if p.val != q.val: return False

        return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)

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