2016年Google面筋记录
记录周围伙伴们面试Google遇到的问题和大致解决思路。
【算法】
ARRAY | 数组
1.给定一个有序数组,和两个值表示区间的开始和结束,找出数组中不在这个区间里的元素
输出[2,2], [3]
3.设计电信运营商选号系统
TREE | 树
1. Binary tree maximum path sum.
问题描述
Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example:
Given the below binary tree,
1 / \ 2 3
Return 6
.
解决思路
递归
1. 得到左子树的最大值,Math.max(root.left, 0),0代表不考虑将左子树的最大值加入;
2. 得到右子树的最大值,Math.max(root.right, 0),0代表不考虑将右子树的最大值加入;
3. 以root为根节点,树的最大值为root.val + Math.max(leftMax, rightMax).
边界条件为root为null时,返回0.
程序
public class Solution { private int max = Integer.MIN_VALUE; public int maxPathSum(TreeNode root) { if (root == null) { return 0; } max = root.val; getSumMaxRec(root); return max; } private int getSumMaxRec(TreeNode root) { if (root == null) { return 0; } int leftMax = Math.max(getSumMaxRec(root.left), 0); int rightMax = Math.max(getSumMaxRec(root.right), 0); max = Math.max(max, root.val + leftMax + rightMax); // update return root.val + Math.max(leftMax, rightMax); } }
2. Interleaving String
问题描述
Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
For example,
Given:
s1 = "aabcc"
,
s2 = "dbbca"
,
When s3 = "aadbbcbcac"
, return true.
When s3 = "aadbbbaccc"
, return false.
解决思路
动态规划
递推式为dp[i+1][j+1]代表s3[0, i + j + 1]能否由s1[0, i]和s2[0, j]交替组成。
dp[i+1][j+1] = (dp[i][j+1] && s3[i+j+1] == s1[i+1]) || (dp[i+1][j] && s3[i+j+1] == s2[j+1]).
程序
public class Solution { public boolean isInterleave(String s1, String s2, String s3) { if (s1 == null || s2 == null || s3 == null) { return false; } int len1 = s1.length(); int len2 = s2.length(); int len = s3.length(); if (len != len1 + len2) { return false; } boolean[][] dp = new boolean[len1 + 1][len2 + 1]; // init dp[0][0] = true; for (int i = 0; i < len1; i++) { dp[i + 1][0] = (s3.charAt(i) == s1.charAt(i) && dp[i][0]); } for (int i = 0; i < len2; i++) { dp[0][i + 1] = (s3.charAt(i) == s2.charAt(i) && dp[0][i]); } for (int i = 0; i < len1; i++) { for (int j = 0; j < len2; j++) { dp[i + 1][j + 1] = (dp[i][j + 1] && s3.charAt(i + j + 1) == s1.charAt(i)) || (dp[i + 1][j] && s3.charAt(i + j + 1) == s2.charAt(j)); } } return dp[len1][len2]; } }