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

比如[1,2,4] 1,10
输出[3,5-10]

2.给两个集合，找出两个集合中的不一样的部分

[1,2,2,2], [1,2,3]
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];
}
}