标签：arraylist for pre binary search com class sub htm
https://leetcode.com/problems/is-subsequence/#/solutions
http://www.cnblogs.com/EdwardLiu/p/6116896.html
public boolean isSubsequence(String s, String t) {
int j = 0, i = 0;
while (j < t.length() && i < s.length()) {
if (s.charAt(i) == t.charAt(j)) {
i++;
j++;
} else {
j++;
}
}
if (i == s.length()) {
return true;
} else {
return false;
}
}
Follow Up:
The best solution is to create a map for String t, key is char, value is the index of appearance in ascending order
public boolean isSubsequence(String s, String t) {
List<Integer>[] idx = new List[256]; // Just for clarity 字典集合, 每一个字母的ASCII 码当作键, 在t中的顺序当作值
for (int i = 0; i < t.length(); i++) { // 生成字典
if (idx[t.charAt(i)] == null)
idx[t.charAt(i)] = new ArrayList<>();
idx[t.charAt(i)].add(i);
}
int prev = 0; // 前一个字母在t中的坐标, 控制升序
for (int i = 0; i < s.length(); i++) { //开始在字典里查找
if (idx[s.charAt(i)] == null) return false; // Note: char of S does NOT exist in T causing NPE
int j = Collections.binarySearch(idx[s.charAt(i)], prev); //在当前字母表中查找其比之前的字母升序的坐标
if (j < 0) j = -j - 1; // 二分搜索的注意点
if (j == idx[s.charAt(i)].size()) return false; //在升序后找不到false
prev = idx[s.charAt(i)].get(j) + 1; //在t中查找当前s的字母比s中的前一个字母在t中升序的坐标
}
return true;
}
标签：arraylist for pre binary search com class sub htm
原文地址：http://www.cnblogs.com/apanda009/p/7101774.html
