约瑟夫环
有100个人，环成一圈开始报数，从1数起，数到7就枪毙一个，然后继续从1数起，数到7就枪毙一个，直到所有枪毙。求所有人的死亡顺序。
解一
使用哈希大法，不断将枪毙的人标记到哈希中
function kill() {
var ret = [], map = {}, j = 0, loopIndex = 0
do {
for (var i = 1; i <= 100; i++) {
if (!map[i]) {
j++
if (j === 7) {
ret.push(i)
map[i] = true
j = 0
}
}
loopIndex++
}
if (ret.length === 100) {//所有人枪毙完
break
}
} while (1);
console.log(ret)
console.log("测试复杂度", loopIndex)
return ret;
}
kill()
好写好有，就是复杂度有点高，外循环跑了34次。但数组又不方便移除，虽然说JS有splice方法。
解二
弄一个链表，然后围起来枪毙，挂掉的踢出圈子。
function kill2() {
var ret = [], j = 0
var head, node
for (var i = 1; i <= 100; i++) {
if (!node) {
node = {
value: i
}
head = node
} else {
node = node.next = {
value: i
}
}
}
node.next = head;
//形成一个环链表
var loopIndex = 0
var prev = node;
var curr = head;
while (curr.next) {
var next = curr.next
loopIndex++
if (curr === next) {
//只剩下自己了
ret.push(curr.value)
break
}
j++
if (j === 7) {//喊号为7的倒霉蛋
curr.next = null
prev.next = next;
ret.push(curr.value)
curr = next
j = 0
continue
}
prev = curr
curr = next
}
console.log(ret)
console.log("loopIndex", loopIndex)
return ret
}
kill2()
解三
在生成链表时，可以求解一部分答案，对应的数就不用变成节点，这样链表也短很多。
function kill3() {
var ret = [], j = 0
var head, node, begin
for (var i = 1; i <= 100; i++) {
if (i % 7 == 0) { //减少链表的生成
ret.push(i)
begin = node;//链表的开始
continue
}
if (!node) {
node = {
value: i
}
head = node
} else {
node = node.next = {
value: i
}
}
}
node.next = head;
//形成一个环链表
var loopIndex = 0
var prev = begin
var curr = prev.next
while (curr.next) {
var next = curr.next
loopIndex++;
if (curr === next) {
ret.push(curr.value)
break
}
j++
if (j === 7) {
curr.next = null
ret.push(curr.value)
prev.next = next;
curr = next
j = 0
continue
}
prev = curr
curr = next
}
console.log(ret)
console.log("loopIndex", loopIndex)
return ret
}
kill3()
发布于 2019-07-05 00:52
