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题目描述:
- 给定一个字符串数组 words,找到 length(word[i]) * length(word[j]) 的最大值,并且这两个单词不含有公共字母。你可以认为每个单词只包含小写字母。如果不存在这样的两个单词,返回 0。
- 示例 1:
- 输入: ["abcw","baz","foo","bar","xtfn","abcdef"]
- 输出: 16
- 解释: 这两个单词为 "abcw", "xtfn"。
- 示例 2:
- 输入: ["a","ab","abc","d","cd","bcd","abcd"]
- 输出: 4
- 解释: 这两个单词为 "ab", "cd"。
- 示例 3:
- 输入: ["a","aa","aaa","aaaa"]
- 输出: 0
- 解释: 不存在这样的两个单词。
- 来源:力扣(LeetCode)
- 链接:https://leetcode-cn.com/problems/maximum-product-of-word-lengths
- 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
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- class Solution {
- public:
- int bitNumber(char ch) {
- return (int)ch - (int)'a';
- }
- int maxProduct(vector<string>& words) {
- int n = words.size();
- vector<int> masks = vector<int>(n);
- vector<int> lens = vector<int>(n);
- //位运算
- int bitmask = 0;
- for (int i = 0; i < n; ++i) {
- bitmask = 0;
- for (char cha : words[i]) {
- bitmask |= 1 << bitNumber(cha);
- }
- masks[i] = bitmask;
- lens[i] = words[i].size();
- }
- int maxVal = 0;
- for (int i = 0; i < n; ++i){
- for (int j = i + 1; j < n; ++j){
- if ((masks[i] & masks[j]) == 0){
- maxVal = max(maxVal, lens[i] * lens[j]);
- }
- }
- }
- return maxVal;
- }
- };
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参考链接:https://leetcode-cn.com/problems ... eng-ji-by-leetcode/ |
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