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merged 2 commits into from
Nov 9, 2021
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✨update: Modify 488 #213

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✨update: Modify 488
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✨update: Modify 488
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18 changes: 12 additions & 6 deletions LeetCode/481-490/488. 祖玛游戏（困难）.md
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Original file line number Diff line number Diff line change
Expand Up @@ -82,7 +82,7 @@ Tag : 「DFS」、「搜索」、「启发式搜索」

但在爆搜过程中同时维持两个字符串构造会超时，考虑使用一个 `int` 来记录 $hand$ 的使用情况。

![image.png](https://pic.leetcode-cn.com/1636422194-vvIBfl-image.png)
![image.png](https://pic.leetcode-cn.com/1636436638-kqDwIl-image.png)

代码：
```Java
Expand All @@ -106,6 +106,10 @@ class Solution {
if (((cur >> i) & 1) == 1) continue;
int next = (1 << i) | cur;
for (int j = 0; j <= n; j++) {
boolean ok = false;
if (j > 0 && j < n && a.charAt(j) == a.charAt(j - 1) && a.charAt(j - 1) != b.charAt(i)) ok = true;
if (j < n && a.charAt(j) == b.charAt(i)) ok = true;
if (!ok) continue;
StringBuilder sb = new StringBuilder();
sb.append(a.substring(0, j)).append(b.substring(i, i + 1));
if (j != n) sb.append(a.substring(j));
Expand Down Expand Up @@ -139,7 +143,7 @@ class Solution {

我们建立一个类 `Node` 来代指当前搜索局面。

```Java
```Java []
class Node {
// 当前的棋盘状况
String a;
Expand Down Expand Up @@ -169,7 +173,7 @@ class Node {

需要注意的是：对于某个局面 $node$ 而言，最终的距离是由「已确定距离」+「估值距离」两部分组成，我们应当根据这两部分之和进行出队，才能确保算法的正确性。

![image.png](https://pic.leetcode-cn.com/1636424014-cCaHWU-image.png)
![image.png](https://pic.leetcode-cn.com/1636436200-gPQhnD-image.png)

代码：
```Java
Expand Down Expand Up @@ -206,9 +210,7 @@ class Solution {
public int findMinStep(String _a, String _b) {
b = _b;
m = b.length();
PriorityQueue<Node> q = new PriorityQueue<>((o1,o2)->{
return (o1.val + o1.step) - (o2.val + o2.step);
});
PriorityQueue<Node> q = new PriorityQueue<>((o1,o2)->(o1.val+o1.step)-(o2.val+o2.step));
q.add(new Node(_a, 1 << m, f(_a, 1 << m), 0));
map.put(_a, 0);
while (!q.isEmpty()) {
Expand All @@ -221,6 +223,10 @@ class Solution {
if (((cur >> i) & 1) == 1) continue;
int next = (1 << i) | cur;
for (int j = 0; j <= n; j++) {
boolean ok = false;
if (j > 0 && j < n && a.charAt(j) == a.charAt(j - 1) && a.charAt(j - 1) != b.charAt(i)) ok = true;
if (j < n && a.charAt(j) == b.charAt(i)) ok = true;
if (!ok) continue;
StringBuilder sb = new StringBuilder();
sb.append(a.substring(0, j)).append(b.substring(i, i + 1));
if (j != n) sb.append(a.substring(j));
Expand Down