* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
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1.9 KiB
id, title, challengeType, videoUrl, dashedName
id | title | challengeType | videoUrl | dashedName |
---|---|---|---|---|
5900f37a1000cf542c50fe8d | 问题14:最长的Collatz序列 | 5 | problem-14-longest-collatz-sequence |
--description--
为正整数集定义以下迭代序列:
n
→ n
/ 2( n
是偶数)
n
→3 n
+ 1( n
为奇数)
使用上面的规则并从13开始,我们生成以下序列:
13→40→20→10→5→16→8→4→2→1
可以看出,该序列(从13开始并在1结束)包含10个项。虽然尚未证实(Collatz问题),但是认为所有起始数字都在1处结束。在给定limit
下,哪个起始数产生最长链?注意:一旦链条启动,条款允许超过一百万。
--hints--
longestCollatzSequence(14)
应该返回9。
assert.strictEqual(longestCollatzSequence(14), 9);
longestCollatzSequence(5847)
应返回3711。
assert.strictEqual(longestCollatzSequence(5847), 3711);
longestCollatzSequence(46500)
应返回35655。
assert.strictEqual(longestCollatzSequence(46500), 35655);
longestCollatzSequence(54512)
应返回52527。
assert.strictEqual(longestCollatzSequence(54512), 52527);
longestCollatzSequence(1000000)
应返回837799。
assert.strictEqual(longestCollatzSequence(100000), 77031);
--seed--
--seed-contents--
function longestCollatzSequence(limit) {
return true;
}
longestCollatzSequence(14);
--solutions--
function longestCollatzSequence(limit) {
let longest = 1;
let maxLength = 1;
for (let i = Math.floor(limit / 2); i < limit; i++) {
let len = colLen(i);
if (len > maxLength) {
longest = i;
maxLength = len;
}
}
return longest;
}
const knownSequence = { '1': 1 };
function colLen(n) {
if (knownSequence[n]) {
return knownSequence[n];
} else {
const len = n % 2 === 0 ? colLen(n / 2) + 1 : colLen((3 * n + 1) / 2) + 2;
knownSequence[n] = len;
return len;
}
}