Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0. Example 1: Given [“abcw”, “baz”...
leetcode | Maximum Product of Word Lengths Given a string arraywords, find the maximum value oflength(word[i]) * length(word[j])where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0. ...
小写字母a-z是26位,一般统计是否存在我们要申请一个bool flg[26]这样的数组,但是我们在这里用int代替,int是32位可以替代flg数组,用与(&),或(1),以及向左移位(<<)就能完成。如“abcd” 的int值为 0000 0000 0000 0000 0000 0000 0000 1111,“wxyz” 的int值为 1111 0000 0000 0000 0000 0000 0000 000...
{ if(ans < words[i].length()*words[j].length()) ans = words[i].length()*words[j].length(); } } } return ans; } int compute(string s) { int bit = 0; for(int i=0;i<s.length();i++) bit |= (1<<(s[i]-'a')); return bit; } }; 1. 2. 3. 4. 5. 6. 7....
Explanation: No such pair of words. 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"] ...
public int maxProduct(String[] words) { int best = 0; //保存每个字符串的二进制表示形式 int[] bitsets = new int[words.length]; // 先循环每个字符串 for (int i = 0; i < words.length; i++) { int wlen = words[i].length(), bitset = 0; ...
Given an integer arraynums, find the contiguous subarray within an array (containing at least one number) which has the largest product. Example 1: <pre class="highlighter-hljs" highlighted="true" has-selection="true" style="margin: 10px auto; padding: 0px; transition-duration: 0.2s; tran...
**解析:**Version 1,两层循环遍历,O(N^2)。 Version 1 代码语言:javascript 复制 classSolution:defmaxProduct(self,nums:List[int])->int:product=0length=len(nums)foriinrange(length):forjinrange(i+1,length):product=max(product,(nums[i]-1)*(nums[j]-1))returnproduct ...
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