这题我通过模仿第一题的做法采用hashmap将时间复杂度降至O(n^2),不过leetcode似乎把时间复杂度降到了O(n),或许这就是大佬写的吧。。。。
我的代码
public int lengthOfLongestSubstring(String s) {
if(s.length() == 0) {
return 0;
}else if(s.length() == 1){
return 1;
}
Map<Integer, Character> map = new HashMap<Integer, Character>();
//int k = 0;
int[] c = new int[s.length()];
if(s.length() == 0) {
return 0;
}
for(int i = 0;i < s.length();i++) {
map.clear();
for(int j = i;j < s.length();j++) {
if(map.containsValue(s.charAt(j)) == false) {
map.put(j, s.charAt(j));
c[i] = j - i + 1 ;
}else {
//c[i] = j - i;
break;
}
}
}
int k = 0;
for(int i = 0;i < c.length;i++) {
if(c[i] > c[k]) {
k = i;
}
}
return c[k];
}
大佬们的
solution 1
public class Solution {
public int lengthOfLongestSubstring(String s) {
int n = s.length(), ans = 0;
Map<Character, Integer> map = new HashMap<>(); // current index of character
// try to extend the range [i, j]
for (int j = 0, i = 0; j < n; j++) {
if (map.containsKey(s.charAt(j))) {
i = Math.max(map.get(s.charAt(j)), i);
}
ans = Math.max(ans, j - i + 1);
map.put(s.charAt(j), j + 1);
}
return ans;
}
}
solution2 通过使用ASCII实现
public class Solution {
public int lengthOfLongestSubstring(String s) {
int n = s.length(), ans = 0;
int[] index = new int[128]; // current index of character
// try to extend the range [i, j]
for (int j = 0, i = 0; j < n; j++) {
i = Math.max(index[s.charAt(j)], i);
ans = Math.max(ans, j - i + 1);
index[s.charAt(j)] = j + 1;
}
return ans;
}
}