@@ -146,9 +146,9 @@ class Solution {
146146 Deque<String > d1 = new ArrayDeque<> (), d2 = new ArrayDeque<> ();
147147 /*
148148 * m1 和 m2 分别记录两个方向出现的状态是经过多少次转换而来
149- * e.g
150- * m1 = {"1000":1} 代表 "1000" 由 s="0000" 替换 1 次字符而来
151- * m2 = {"9999":3} 代表 "9999" 由 t="9996" 替换 3 次字符而来
149+ * e.g.
150+ * m1 = {"1000":1} 代表 "1000" 由 s="0000" 旋转 1 次而来
151+ * m2 = {"9999":3} 代表 "9999" 由 t="9996" 旋转 3 次而来
152152 */
153153 Map<String , Integer > m1 = new HashMap<> (), m2 = new HashMap<> ();
154154 d1. addLast(s);
@@ -174,33 +174,36 @@ class Solution {
174174 return - 1 ;
175175 }
176176 int update (Deque<String > deque , Map<String , Integer > cur , Map<String , Integer > other ) {
177- String poll = deque. pollFirst();
178- char [] pcs = poll. toCharArray();
179- int step = cur. get(poll);
180- // 枚举替换哪个字符
181- for (int i = 0 ; i < 4 ; i++ ) {
182- // 能「正向转」也能「反向转」,这里直接枚举偏移量 [-1,1] 然后跳过 0
183- for (int j = - 1 ; j <= 1 ; j++ ) {
184- if (j == 0 ) continue ;
177+ int m = deque. size();
178+ while (m-- > 0 ) {
179+ String poll = deque. pollFirst();
180+ char [] pcs = poll. toCharArray();
181+ int step = cur. get(poll);
182+ // 枚举替换哪个字符
183+ for (int i = 0 ; i < 4 ; i++ ) {
184+ // 能「正向转」也能「反向转」,这里直接枚举偏移量 [-1,1] 然后跳过 0
185+ for (int j = - 1 ; j <= 1 ; j++ ) {
186+ if (j == 0 ) continue ;
185187
186- // 求得替换字符串 str
187- int origin = pcs[i] - ' 0'
188- int next = (origin + j) % 10 ;
189- if (next == - 1 ) next = 9 ;
188+ // 求得替换字符串 str
189+ int origin = pcs[i] - ' 0'
190+ int next = (origin + j) % 10 ;
191+ if (next == - 1 ) next = 9 ;
190192
191- char [] clone = pcs. clone();
192- clone[i] = (char )(next + ' 0'
193- String str = String . valueOf(clone);
193+ char [] clone = pcs. clone();
194+ clone[i] = (char )(next + ' 0'
195+ String str = String . valueOf(clone);
194196
195- if (set. contains(str)) continue ;
196- if (cur. containsKey(str)) continue ;
197- 198- // 如果在「另一方向」找到过,说明找到了最短路,否则加入队列
199- if (other. containsKey(str)) {
200- return step + 1 + other. get(str);
201- } else {
202- deque. addLast(str);
203- cur. put(str, step + 1 );
197+ if (set. contains(str)) continue ;
198+ if (cur. containsKey(str)) continue ;
199+ 200+ // 如果在「另一方向」找到过,说明找到了最短路,否则加入队列
201+ if (other. containsKey(str)) {
202+ return step + 1 + other. get(str);
203+ } else {
204+ deque. addLast(str);
205+ cur. put(str, step + 1 );
206+ }
204207 }
205208 }
206209 }
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