sovle #139

Author: aylei

src/lib.rs

@@ -135,3 +135,4 @@ mod n0134_gas_station;
 mod n0135_candy;
 mod n0136_single_number;
 mod n0137_single_number_ii;
+mod n0139_word_break;

src/n0139_word_break.rs

@@ -0,0 +1,81 @@
+/**
+ * [139] Word Break
+ *
+ * Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
+ * 
+ * Note:
+ * 
+ * 
+ * 	The same word in the dictionary may be reused multiple times in the segmentation.
+ * 	You may assume the dictionary does not contain duplicate words.
+ * 
+ * 
+ * Example 1:
+ * 
+ * 
+ * Input: s = "leetcode", wordDict = ["leet", "code"]
+ * Output: true
+ * Explanation: Return true because "leetcode" can be segmented as "leet code".
+ * 
+ * 
+ * Example 2:
+ * 
+ * 
+ * Input: s = "applepenapple", wordDict = ["apple", "pen"]
+ * Output: true
+ * Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
+ *              Note that you are allowed to reuse a dictionary word.
+ * 
+ * 
+ * Example 3:
+ * 
+ * 
+ * Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
+ * Output: false
+ * 
+ * 
+ */
+pub struct Solution {}
+
+/*
+ 记 f[n] 表示从 0 开始长度为 n 的 substring 是否可以被组成，那么：
+
+ f[n] = f[k] && (s[k..n] in dict)
+ f[0] = true
+
+ DP 向上递推即可
+
+ BFS 也是可以的
+ */
+
+// submission codes start here
+
+use std::collections::HashSet;
+impl Solution {
+    pub fn word_break(s: String, word_dict: Vec<String>) -> bool {
+        let word_dict = word_dict.into_iter().collect::<HashSet<_>>();
+        let mut dp = vec![false; s.len()+1];
+        dp[0] = true;
+        for i in 1..s.len()+1 {
+            for j in 0..s.len() {
+                if dp[j] && word_dict.contains(&s[j..i]) {
+                    dp[i] = true;
+                }
+            }
+        }
+        dp[s.len()]
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_139() {
+        assert_eq!(Solution::word_break("leetcode".to_owned(), vec_string!["leet", "code"]), true);
+        assert_eq!(Solution::word_break("catsandog".to_owned(), vec_string!["cats", "dog", "sand", "and", "cat"]), false);
+    }
+}