代码随想录算法训练营day37| 738. 单调递增的数字、968. 监控二叉树、贪心算法总结

738、单调递增的数字：

class Solution(object):
    def monotoneIncreasingDigits(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n == 0:
            return 0
        nums = [int(i) for i in str(n)]
        flag = len(nums)
        for i in range(len(nums)-1, 0, -1):
            if nums[i] < nums[i-1]:
                nums[i-1] -= 1
                flag = i
        for i in range(len(nums)):
            if i >= flag:
                nums[i] = 9
        return int(''.join([str(i) for i in nums]).lstrip('0'))

选择从前往后遍历还是从后往前遍历是本题的关键，一旦确定了从后往前遍历，注意代码技巧就可以了

968、监控二叉树：

class Solution(object):
    def __init__(self):
        self.res = 0

    def traversal(self, cur):
        if not cur:
            return 2
        left = self.traversal(cur.left)
        right = self.traversal(cur.right)
        if left == 2 and right == 2:
            return 0
        if left == 0 or right == 0:
            self.res += 1
            return 1
        if left == 1 or right == 1:
            return 2
        return -1

    def minCameraCover(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if self.traversal(root) == 0:
            self.res += 1
        return self.res

本题是贪心算法和二叉树的结合，需要约定好不同情形下左右子树传回的值，来确定当前节点的状态，从而判断监控的数量