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package class_2022_12_4_week;
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import java.util.HashSet;
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import java.util.PriorityQueue;
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// 对于某些非负整数 k ，如果交换 s1 中两个字母的位置恰好 k 次
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// 能够使结果字符串等于 s2 ，则认为字符串 s1 和 s2 的 相似度为 k
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// 给你两个字母异位词 s1 和 s2 ，返回 s1 和 s2 的相似度 k 的最小值
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// 测试链接 : https://leetcode.cn/problems/k-similar-strings/
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public class Code06_KSimilarStrings {
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public static class Node {
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public int cost; // 代价，已经换了几回了！
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public int guess;// 猜测还要换几回，能变对！
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public int where;// 有必须去比对的下标，左边不再换了！
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public String str; // 当前的字符
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public Node(int r, int g, int i, String s) {
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cost = r;
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guess = g;
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where = i;
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str = s;
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}
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}
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public static int kSimilarity(String s1, String s2) {
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int len = s1.length();
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PriorityQueue<Node> heap = new PriorityQueue<>((a, b) -> (a.cost + a.guess) - (b.cost + b.guess));
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HashSet<String> visited = new HashSet<>();
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heap.add(new Node(0, 0, 0, s1));
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while (!heap.isEmpty()) {
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Node cur = heap.poll();
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if (visited.contains(cur.str)) {
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continue;
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}
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if (cur.str.equals(s2)) {
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return cur.cost;
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}
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visited.add(cur.str);
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int firstDiff = cur.where;
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while (cur.str.charAt(firstDiff) == s2.charAt(firstDiff)) {
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firstDiff++;
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}
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char[] curStr = cur.str.toCharArray();
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for (int i = firstDiff + 1; i < len; i++) {
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if (curStr[i] == s2.charAt(firstDiff) && curStr[i] != s2.charAt(i)) {
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swap(curStr, firstDiff, i);
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add(String.valueOf(curStr), s2, cur.cost + 1, firstDiff + 1, heap, visited);
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swap(curStr, firstDiff, i);
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}
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}
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}
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return -1;
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}
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public static void swap(char[] s, int i, int j) {
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char tmp = s[i];
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s[i] = s[j];
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s[j] = tmp;
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}
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public static void add(String add, String s2, int cost, int index, PriorityQueue<Node> heap,
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HashSet<String> visited) {
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if (!visited.contains(add)) {
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heap.add(new Node(cost, evaluate(add, s2, index), index, add));
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}
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}
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// 估值函数
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// 看每周有营养的大厂算法面试题，2022年1月第3周
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// 估值函数的估计值要绝对 <= 真实距离
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// 但又要确保估计值足够大足够接近真实距离，这样效果最好
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public static int evaluate(String s1, String s2, int index) {
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int diff = 0;
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for (int i = index; i < s1.length(); i++) {
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if (s1.charAt(i) != s2.charAt(i)) {
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diff++;
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}
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}
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return (diff + 1) / 2;
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}
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}
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