比赛地址
https://leetcode.com/contest/weekly-contest-81
821. Shortest Distance to a Character
Given a string S and a character C, return an array of integers representing the shortest distance from the character C in the string.
Example 1:
Input: S = "loveleetcode", C = 'e' Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]
找出每个位置左边右边离C字母最近的距离，简单难度，两次遍历分别求出左边和右边最近距离即可。
/**
* @param {string} S
* @param {character} C
* @return {number[]}
*/
var shortestToChar = function(S, C) {
var r = new Array(S.length).fill(99999);
var pos = -1;
for (var i = S.length - 1; i >= 0; i--) {
if (S[i] == C) {
pos = i;
}
if (pos >= 0) {
r[i] = pos - i;
}
}
pos = -1;
for (var i = 0; i < S.length; i++) {
if (S[i] == C) {
pos = i;
}
if (pos >= 0) {
r[i] = Math.min(r[i], i - pos);
}
}
return r;
};
822. Card Flipping Game
On a table are N cards, with a positive integer printed on the front and back of each card (possibly different).
We flip any number of cards, and after we choose one card.
If the number X on the back of the chosen card is not on the front of any card, then this number X is good.
What is the smallest number that is good? If no number is good, output 0.
Here, fronts[i] and backs[i] represent the number on the front and back of card i.
A flip swaps the front and back numbers, so the value on the front is now on the back and vice versa.
Example:
Input: fronts = [1,2,4,4,7], backs = [1,3,4,1,3] Output:2Explanation: If we flip the second card, the fronts are[1,3,4,4,7]and the backs are[1,2,4,1,3]. We choose the second card, which has number 2 on the back, and it isn't on the front of any card, so2is good.
一大片英文不大看得懂，纯属坑中国人的题，试错法试了四次看错误case猜题目意思、大致意思就是牌随便翻转，找出翻了之后最小的只有一张的牌。
/**
* @param {number[]} fronts
* @param {number[]} backs
* @return {number}
*/
var flipgame = function(fronts, backs) {
var f = new Array(2001).fill(false);
for (var i = 0; i < fronts.length; i++) {
if (fronts[i] == backs[i]) {
f[fronts[i]] = true;
}
}
var ans = 2001;
for (var i = 0; i < fronts.length; i++) {
if (fronts[i] !== backs[i]) {
if (!f[fronts[i]]) {
ans = Math.min(ans, fronts[i]);
}
if (!f[backs[i]]) {
ans = Math.min(ans, backs[i]);
}
}
}
return ans === 2001 ? 0 : ans;
};
820. Short Encoding of Words
Given a list of words, we may encode it by writing a reference string S and a list of indexes A.
For example, if the list of words is ["time", "me", "bell"], we can write it as S = "time#bell#" and indexes = [0, 2, 5].
Then for each index, we will recover the word by reading from the reference string from that index until we reach a "#" character.
What is the length of the shortest reference string S possible that encodes the given words?
Example:
Input: words =["time", "me", "bell"]Output: 10 Explanation: S ="time#bell#" and indexes = [0, 2, 5].
题意是压缩所有单词到尽量短的一句话中，使得所有单词都可以用一个位置表示。
思路是后缀树。Js其实有一种很方便的树的实现例如本题单词time，生成这样一个后缀树结构：{e:{m:{i:{t:{}}}}}，当遇到下个单词如me时，从最外层开始找到{e:{m:{...}}}说明单词me已经被包含。具体代码如下：
/**
* @param {string[]} words
* @return {number}
*/
var minimumLengthEncoding = function(words) {
words.sort((a,b)=>b.length-a.length);
var ans = 0;
var d = {};
for (var i = 0; i < words.length; i++) {
var find = true;
var p = d;
for (var j = words[i].length - 1; j >= 0; j--) {
if (!p[words[i][j]]) {
find = false;
p[words[i][j]] = {};
}
p = p[words[i][j]];
}
if (!find) {
ans += words[i].length + 1;
}
}
return ans;
};
823. Binary Trees With Factors
Given an array of unique integers, each integer is strictly greater than 1.
We make a binary tree using these integers and each number may be used for any number of times.
Each non-leaf node's value should be equal to the product of the values of it's children.
How many binary trees can we make? Return the answer modulo 10 ** 9 + 7.
Example 1:
Input:A = [2, 4]Output: 3 Explanation: We can make these trees:[2], [4], [4, 2, 2]
Example 2:
Input:A = [2, 4, 5, 10]Output:7Explanation: We can make these trees:[2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2].
题意是求满足这样规则的树的数量：非叶结点的值是两个孩子节点的乘积。
中等难度、就是计算结果要求对1000000007取模，马蛋又被坑了一次。
利用集合加速查找数字是否存在、键值对m记录中间结果。
/**
* @param {number[]} A
* @return {number}
*/
var numFactoredBinaryTrees = function(A) {
var s = new Set(A);
var m = {};
var M = 1000000007;
A.sort((a,b)=>a-b);
for (var i = 0; i < A.length; i++) {
var count = 1;
for (var j = 0; j < i; j++) {
if (A[i] % A[j] === 0) {
var other = A[i] / A[j];
if (s.has(other)) {
count = (count + (m[A[j]] % M) * (m[other] % M)) % M;
}
}
}
m[A[i]] = count;
}
var ans = 0;
for (var k in m) {
ans = (ans + m[k]) % M;
}
return ans;
};