标签:return String int Explanation two BitMask words 相关
318. Maximum Product of Word LengthsGiven a string array words
, return the maximum value of length(word[i]) * length(word[j])
where the two words do not share common letters. If no such two words exist, return 0
.
Example 1:
Input: words = ["abcw","baz","foo","bar","xtfn","abcdef"] Output: 16 Explanation: The two words can be "abcw", "xtfn".
Example 2:
Input: words = ["a","ab","abc","d","cd","bcd","abcd"] Output: 4 Explanation: The two words can be "ab", "cd".
Example 3:
Input: words = ["a","aa","aaa","aaaa"] Output: 0 Explanation: No such pair of words.
Constraints:
2 <= words.length <= 1000
1 <= words[i].length <= 1000
words[i]
consists only of lowercase English letters.
暴力解法
class Solution { public int maxProduct(String[] words) { int max = 0; for(int i=0;i<words.length-1;i++){ for(int j=i+1;j<words.length;j++){ if(!hasCommon(words[i],words[j])) max = Math.max(max,words[i].length()*words[j].length()); } } return max; } private boolean hasCommon(String w1,String w2){ for(char c : w1.toCharArray()){ if(w2.indexOf(c)>=0) return true; } return false; } }
巧用bitmask,将 字符串比对 过程从O(len2) 到 O(1), 最终时间复杂度: O(N*N+L) N为字符串个数,L为字符串长度
class Solution { public int maxProduct(String[] words) { int max = 0; int[] state = new int[words.length]; for(int i=0;i<words.length;i++){ state[i] = getState(words[i]); } for(int i=0;i<words.length-1;i++){ for(int j=i+1;j<words.length;j++){ if( (state[i] & state[j]) ==0 ) max = Math.max(max,words[i].length()*words[j].length()); } } return max; } private int getState(String str){ int mask = 0; for(char c:str.toCharArray()){ int pos = c-'a'; mask |= 1<<(pos); } return mask; } }
bitmask 基础上增加 map对重复字符串集合进行去重,减少比对次数
class Solution { public int maxProduct(String[] words) { int max = 0; Map<Integer,Integer> map = new HashMap(); for(int i=0;i<words.length;i++){ int bitValue = getState(words[i]); map.put(bitValue, Math.max(map.getOrDefault(bitValue,0), words[i].length())); } int result = 0; for(int i:map.keySet()){ for(int j:map.keySet()){ if( i!=j && ( (i & j) == 0)) result = Math.max(result, map.get(i)*map.get(j)); } } return result; } private int getState(String str){ int mask = 0; for(char c:str.toCharArray()){ int pos = c-'a'; mask |= 1<<(pos); } return mask; } }
标签:return,String,int,Explanation,two,BitMask,words,相关 来源: https://www.cnblogs.com/cynrjy/p/16325459.html
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