Day 37 | 738. Monotone Increasing Digits | 714. Best Time to Buy and Sell Stock | 968. BT Camera

原创已于 2023-01-19 15:10:32 修改 · 200 阅读

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于 2023-01-19 14:26:04 首次发布

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该文详细记录了在LeetCode上的一系列算法题解，涉及股票的最大利润计算、保证单调递增数字的构建以及最小摄像头覆盖的二叉树问题。通过这些问题，展示了对数组、链表、二叉树结构的处理策略以及动态规划和深度优先搜索的应用。

LeetCode 738. Monotone Increasing Digits

Question Link

class Solution {
    public int monotoneIncreasingDigits(int n) {
        String  s = String.valueOf(n);
        char[] chars = s.toCharArray();
        int start = s.length();
        for(int i = s.length() - 2; i >= 0; i--){
            if(chars[i] > chars[i+1]){
                chars[i]--;
                start = i + 1;
            }
        }

        for(int i = start; i < s.length(); i++)
            chars[i] = '9';
        
        return Integer.parseInt(String.valueOf(chars));
    }
}

When meeting chars[i - 1] > chars[i], let chars[i-1]-- and set chars[i] as 9. That can ensure these two digits become the largest monotone-increasing integers
The traverse order is from back to front. Because if we traverse from front to back, we can’t reuse the result of the last compare

LeetCode 714. Best Time to Buy and Sell Stock with Transaction Fee

Question Link

class Solution {
    public int maxProfit(int[] prices, int fee) {
        int buy = prices[0] + fee;
        int sum = 0;
        for(int p : prices){
            if(p + fee < buy)
                buy = p + fee;
            
            else if(p > buy){
                sum += p - buy;
                buy = p;
            }
        }
        return sum;
    }
}

LeetCode 968. BT Camera

Question Link

class Solution {
    int res = 0;
    public int minCameraCover(TreeNode root) {
        if(traverse(root) == 0)
            res++;
        return res;
    }

    int traverse(TreeNode root){
        if(root == null)
            return 2;   // By default, empty nodes have coverage to avoid install cameras on the leaf
        int left = traverse(root.left);
        int right = traverse(root.right);

        if(left == 2 && right == 2)
            return 0;
        else if(left == 0 || right == 0){
            res++;
            return 1;
        }else
            return 2;
    }   
}

The status of nodes:
- 0 means no coverage
- 1 means the node has the camera
- 2 means the node has coverage
By default, empty nodes have coverage to avoid installing cameras on the leaf