leetcode 159. Longest Substring with At Most Two Distinct Characters 求两个字母组成的最大子串长度 --------- jav
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Given a string, find the length of the longest substring T that contains at most 2 distinct characters.
For example, Given s = “eceba”
,
T is "ece" which its length is 3.
给一个字符串,求这个字符串中,由两个字母组成的,连续的最长子串的长度。
虽然是hard,但是感觉并没有什么难度。
用ch1和preCh记录当前两个字母,preCh记录的是上一个字母(s.charAt(i-1)),same记录的是preCh这个字母重复出现的次数,这样出现第三个字母的时候,就可以直接得出从0到i由后两个字母组成的长度为same+1,并没有使用其他的数据结构。
时间O(n),空间O(1).
public class Solution { public int lengthOfLongestSubstringTwoDistinct(String s) { int len = s.length(); if (len < 3){ return len; } char ch1 = s.charAt(0); int same = 0; while (same < len && s.charAt(same) == ch1){ same++; } if (same == len){ return len; } char preCh = s.charAt(same); int result = same + 1; int ans = same + 1; int i = same + 1; same = 1; for (; i < len; i++){ if (s.charAt(i) == preCh){ result++; same++; } else if (s.charAt(i) == ch1){ result++; same = 1; ch1 = preCh; preCh = s.charAt(i); } else { ch1 = preCh; preCh = s.charAt(i); ans = Math.max(ans, result); result = same + 1; same = 1; } } ans = Math.max(ans, result); return ans; } }
2、还可以利用hashMap来做:(参考discuss)
Map数目小于3的时候将字母和他的位置加入Map中。
如果是大于等于3,那么找出距离位置hi最远的一个字母(leftMost),删掉,从leftMost的下一个字母开始到当前位置hi就是当前两个字母的长度。
public class Solution { public int lengthOfLongestSubstringTwoDistinct(String s) { if(s.length() < 1) return 0; HashMap<Character,Integer> index = new HashMap<Character,Integer>(); int lo = 0; int hi = 0; int maxLength = 0; while(hi < s.length()) { if(index.size() <= 2) { char c = s.charAt(hi); index.put(c, hi); hi++; } if(index.size() > 2) { int leftMost = s.length(); for(int i : index.values()) { leftMost = Math.min(leftMost,i); } char c = s.charAt(leftMost); index.remove(c); lo = leftMost+1; } maxLength = Math.max(maxLength, hi-lo); } return maxLength; } }
3、其实不算算法,就是把s先转换成char[]。这样就会达到最快。
public class Solution { public int lengthOfLongestSubstringTwoDistinct(String s) { int len = s.length(); if (len < 3){ return len; } char[] words = s.toCharArray(); char ch1 = words[0]; int i = 1; while (i < len && words[i] == ch1){ i++; } if (i == len){ return len; } int same = 1; char preCh = words[i]; int ans = i + 1; int result = ans; i++; while (i < len){ if (words[i] == preCh){ result++; same++; } else if (words[i] == ch1){ result++; same = 1; ch1 = preCh; preCh = words[i]; } else { ch1 = preCh; preCh = words[i]; ans = Math.max(ans, result); result = same + 1; same = 1; } i++; } ans = Math.max(ans, result); return ans; } }
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