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Detailed explanation of several non-recursive full permutation algorithm code examples in JavaScript
Detailed explanation of several non-recursive full permutation algorithm code examples in JavaScript
回溯（非递归）
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Full Permutation(Non-recursive Backtrack) - Mengliao Software</title>
</head>
<body>
<p>
Full Permutation(Non-recursive Backtrack)<br />
Mengliao Software Studio - Bosun Network Co., Ltd.<br />
2012.03.29</p>
<script type="text/javascript">
/*
全排列（非递归回溯）算法
1、建立位置数组，即对位置进行排列，排列成功后转换为元素的排列；
2、第n个位置搜索方式与八皇后问题类似。
*/
var count = 0;
function show(arr) {
document.write("P<sub>" + ++count + "</sub>: " + arr + "<br />");
}
function seek(index, n) {
var flag = false, m = n; //flag为找到位置排列的标志，m保存正在搜索哪个位置
do {
index[n]++;
if (index[n] == index.length) //已无位置可用
index[n--] = -1; //重置当前位置，回退到上一个位置
else if (!(function () {
for (var i = 0; i < n; i++)
if (index[i] == index[n]) return true;
return false;
})()) //该位置未被选择
if (m == n) //当前位置搜索完成
flag = true;
else
n++;
} while (!flag && n >= 0)
return flag;
}
function perm(arr) {
var index = new Array(arr.length);
for (var i = 0; i < index.length; i++)
index[i] = -1;
for (i = 0; i < index.length - 1; i++)
seek(index, i);
while (seek(index, index.length - 1)) {
var temp = [];
for (i = 0; i < index.length; i++)
temp.push(arr[index[i]]);
show(temp);
}
}
perm(["e1", "e2", "e3", "e4"]);
</script>
</body>
</html>排序（非递归）
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Full Permutation(Non-recursive Sort) - Mengliao Software</title>
</head>
<body>
<p>
Full Permutation(Non-recursive Sort)<br />
Mengliao Software Studio - Bosun Network Co., Ltd.<br />
2012.03.30</p>
<script type="text/javascript">
/*
全排列（非递归求顺序）算法
1、建立位置数组，即对位置进行排列，排列成功后转换为元素的排列；
2、按如下算法求全排列：
设P是1～n(位置编号)的一个全排列：p = p1,p2...pn = p1,p2...pj-1,pj,pj+1...pk-1,pk,pk+1...pn
(1)从排列的尾部开始，找出第一个比右边位置编号小的索引j（j从首部开始计算），即j = max{i | pi < pi+1}
(2)在pj的右边的位置编号中，找出所有比pj大的位置编号中最小的位置编号的索引k，即 k = max{i | pi > pj}
pj右边的位置编号是从右至左递增的，因此k是所有大于pj的位置编号中索引最大的
(3)交换pj与pk
(4)再将pj+1...pk-1,pk,pk+1...pn翻转得到排列p' = p1,p2...pj-1,pj,pn...pk+1,pk,pk-1...pj+1
(5)p'便是排列p的下一个排列
例如：
24310是位置编号0～4的一个排列，求它下一个排列的步骤如下：
(1)从右至左找出排列中第一个比右边数字小的数字2；
(2)在该数字后的数字中找出比2大的数中最小的一个3；
(3)将2与3交换得到34210；
(4)将原来2（当前3）后面的所有数字翻转，即翻转4210，得30124；
(5)求得24310的下一个排列为30124。
*/
var count = 0;
function show(arr) {
document.write("P<sub>" + ++count + "</sub>: " + arr + "<br />");
}
function swap(arr, i, j) {
var t = arr[i];
arr[i] = arr[j];
arr[j] = t;
}
function sort(index) {
for (var j = index.length - 2; j >= 0 && index[j] > index[j + 1]; j--)
; //本循环从位置数组的末尾开始，找到第一个左边小于右边的位置，即j
if (j < 0) return false; //已完成全部排列
for (var k = index.length - 1; index[k] < index[j]; k--)
; //本循环从位置数组的末尾开始，找到比j位置大的位置中最小的，即k
swap(index, j, k);
for (j = j + 1, k = index.length - 1; j < k; j++, k--)
swap(index, j, k); //本循环翻转j+1到末尾的所有位置
return true;
}
function perm(arr) {
var index = new Array(arr.length);
for (var i = 0; i < index.length; i++)
index[i] = i;
do {
var temp = [];
for (i = 0; i < index.length; i++)
temp.push(arr[index[i]]);
show(temp);
} while (sort(index));
}
perm(["e1", "e2", "e3", "e4"]);
</script>
</body>
</html>求模（非递归）
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Full Permutation(Non-recursive Modulo) - Mengliao Software</title>
</head>
<body>
<p>Full Permutation(Non-recursive Modulo)<br />
Mengliao Software Studio - Bosun Network Co., Ltd.<br />
2012.03.29</p>
<script type="text/javascript">
/*
全排列（非递归求模）算法
1、初始化存放全排列结果的数组result，与原数组的元素个数相等；
2、计算n个元素全排列的总数，即n!；
3、从>=0的任意整数开始循环n!次，每次累加1，记为index；
4、取第1个元素arr[0]，求1进制的表达最低位，即求index模1的值w，将第1个元素（arr[0]）插入result的w位置，并将index迭代为index\1；
5、取第2个元素arr[1]，求2进制的表达最低位，即求index模2的值w，将第2个元素（arr[1]）插入result的w位置，并将index迭代为index\2；
6、取第3个元素arr[2]，求3进制的表达最低位，即求index模3的值w，将第3个元素（arr[2]）插入result的w位置，并将index迭代为index\3；
7、……
8、直到取最后一个元素arr[arr.length-1]，此时求得一个排列；
9、当index循环完成，便求得所有排列。
例：
求4个元素["a", "b", "c", "d"]的全排列, 共循环4!=24次，可从任意>=0的整数index开始循环，每次累加1，直到循环完index+23后结束；
假设index=13（或13+24，13+2*24，13+3*24…），因为共4个元素，故迭代4次，则得到的这一个排列的过程为：
第1次迭代，13/1，商=13，余数=0，故第1个元素插入第0个位置（即下标为0），得["a"]；
第2次迭代，13/2, 商=6，余数=1，故第2个元素插入第1个位置（即下标为1），得["a", "b"]；
第3次迭代，6/3, 商=2，余数=0，故第3个元素插入第0个位置（即下标为0），得["c", "a", "b"]；
第4次迭代，2/4，商=0，余数=2, 故第4个元素插入第2个位置（即下标为2），得["c", "a", "d", "b"]；
*/
var count = 0;
function show(arr) {
document.write("P<sub>" + ++count + "</sub>: " + arr + "<br />");
}
function perm(arr) {
var result = new Array(arr.length);
var fac = 1;
for (var i = 2; i <= arr.length; i++)
fac *= i;
for (index = 0; index < fac; index++) {
var t = index;
for (i = 1; i <= arr.length; i++) {
var w = t % i;
for (j = i - 1; j > w; j--)
result[j] = result[j - 1];
result[w] = arr[i - 1];
t = Math.floor(t / i);
}
show(result);
}
}
perm(["e1", "e2", "e3", "e4"]);
</script>
</body>
</html>The above is the detailed content of Detailed explanation of several non-recursive full permutation algorithm code examples in JavaScript. For more information, please follow other related articles on the PHP Chinese website!
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