@@ -251,30 +251,34 @@ class Solution {
251251
252252// 回溯法+记忆化
253253class Solution {
254+ private Set<String > set;
255+ private int [] memo;
254256 public boolean wordBreak (String s , List<String > wordDict ) {
255- Set< String > wordDictSet = new HashSet (wordDict) ;
256- int [] memory = new int [s . length()] ;
257- return backTrack (s, wordDictSet, 0 , memory );
257+ memo = new int [s . length()] ;
258+ set = new HashSet<> (wordDict) ;
259+ return backtracking (s, 0 );
258260 }
259-
260- public boolean backTrack (String s , Set< String > wordDictSet , int startIndex , int [] memory ) {
261- // 结束条件
262- if (startIndex > =. length()) {
261+ 262+ public boolean backtracking (String s , int startIndex ) {
263+ // System.out.println(startIndex);
264+ if (startIndex = =. length()) {
263265 return true ;
264266 }
265- if (memory[startIndex] != 0 ) {
266- // 此处认为:memory[i] = 1 表示可以拼出i 及以后的字符子串, memory[i] = -1 表示不能
267- return memory[startIndex] == 1 ? true : false ;
267+ if (memo[startIndex] == - 1 ) {
268+ return false ;
268269 }
269- for ( int i = startIndex; i < s . length(); ++ i) {
270- // 处理 递归 回溯 循环不变量:[startIndex, i + 1)
271- String word = s. substring(startIndex, i + 1 );
272- if (wordDictSet . contains(word) && backTrack(s, wordDictSet, i + 1 , memory)) {
273- memory[startIndex] = 1 ;
274- return true ;
270+ 271+ for ( int i = startIndex; i < s . length(); i ++ ) {
272+ String sub = s. substring(startIndex, i + 1 );
273+ // 拆分出来的单词无法匹配
274+ if ( ! set . contains(sub)) {
275+ continue ;
275276 }
277+ boolean res = backtracking(s, i + 1 );
278+ if (res) return true ;
276279 }
277- memory[startIndex] = - 1 ;
280+ // 这里是关键,找遍了startIndex~s.length()也没能完全匹配,标记从startIndex开始不能找到
281+ memo[startIndex] = - 1 ;
278282 return false ;
279283 }
280284}
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