Leetcode刷题--Week 1

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前言

眼瞅今年暑假就要开始找工作了，正好借助Leetcode刷题复习一下相关算法，随便为找工作做些准备。接下来每周至少回顾一个算法同时刷一道题，每周打卡。

本周刷题

树

题目：
Given a binary tree, find its minimum depth.The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

解题思路：
思路一：深度优先遍历，递归遍历左右子树，遇到叶子节点时返回 1，返回过程中，比较左右子树，较小深度逐层加1。

public int run(TreeNode root) {
		if(root==null) {
			return 0;
		}
		if(root.left==null && root.right==null) {
			return 1;
		}
		int l=run(root.left);
		int r=run(root.right);
		if(l==0||r==0){ return 1+r+l;}
		return 1+Math.min(l,r);
	}

思路二：广度优先遍历，逐层遍历节点，遇到第一个叶子节点返回其深度；

public int runDFS(TreeNode root){
		if (root == null){return 0;}
		LinkedList<TreeNode> linkedList = new LinkedList<>();
		root.val = 1;
		linkedList.add(root);
		while (!linkedList.isEmpty()){
			TreeNode node = linkedList.pollFirst();
			if (node.left == null && node.right == null){return node.val;}
			if (node.left != null){
				node.left.val = node.val + 1;
				linkedList.add(node.left);
			}
			if (node.right != null ){
				node.right.val = node.val + 1;
				linkedList.add(node.right);
			}
		}
		return  0;

	}

代码：https://github.com/ZhenxueXu/leetcode/blob/master/src/solution/MinimumDepthOfBinaryTree.java

栈

题目：
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are+,-,*,/. Each operand may be an integer or another expression.

Some examples:

["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

解题思路：
根据栈的原理，将数字依次入栈，遇到操作符时，从栈顶弹出两个数字，计算后将结果继续入栈；

public int evalRPN(String[] tokens) {
   	if (tokens == null || tokens.length == 0){
   		return 0;
   	}
   	Stack<Integer> value = new Stack<>();
   	int left = 0;
   	int right = 0;
   	int  result = 0;
   	for (String token : tokens) {
   		switch (token){
   			case "+":
   				right = value.pop();
   				left = value.pop();
   				result = left + right;
   				value.push(result);
   				break;
   			case "-":
   				right = value.pop();
   				left = value.pop();
   				result = left - right;
   				value.push(result);
   				break;
   			case "*":
   				right = value.pop();
   				left = value.pop();
   				result = left * right;
   				value.push(result);
   				break;
   			case "/":
   				right = value.pop();
   				left = value.pop();
   				result = left / right;
   				value.push(result);
   				break;
   				default:
   					value.push(Integer.valueOf(token));
   		}
   	}
   	return value.pop();
   }

代码：https://github.com/ZhenxueXu/leetcode/blob/master/src/solution/EvaluteReversePolishNotation.java