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3 Longest Substring Without Repeating Characters
11 Container With Most Water
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42 Trapping Rain Water
76 Minimum Window Substring
80 Remove Duplicates from Sorted Array II
88 Merge Sorted Array
125 Valid Palindrome
159 Longest Substring with At Most Two Distinct Characters
167 Two Sum II - Input array is sorted
202 Happy Number
209 Minimum Size Subarray Sum
259 3Sum Smaller
283 Move Zeroes
340 Longest Substring with At Most K Distinct Characters
344 Reverse String
345 Reverse Vowels of a String
349 Intersection of Two Arrays
350 Intersection of Two Arrays II
360 Sort Transformed Array
395 Longest Substring with At Least K Repeating Characters
424 Longest Repeating Character Replacement
438 Find All Anagrams in a String
487 Max Consecutive Ones II
524 Longest Word in Dictionary through Deleting
532 K-diff Pairs in an Array
567 Permutation in String
611 Valid Triangle Number
632 Smallest Range
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524 Longest Word in Dictionary through Deleting

1. Question

Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string.
Example 1:
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Input: s = "abpcplea", d = ["ale","apple","monkey","plea"]
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​
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​
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Output: "apple"
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Example 2:
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Input: s = "abpcplea", d = ["a","b","c"]
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​
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Output: "a"
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Note:
  1. 1.
    All the strings in the input will only contain lower-case letters.
  2. 2.
    The size of the dictionary won't exceed 1,000.
  3. 3.
    The length of all the strings in the input won't exceed 1,000.

2. Implementation

(1) Two Pointers
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class Solution {
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public String findLongestWord(String s, List<String> d) {
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if (s == null || s.length() == 0 || d.size() == 0) {
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return "";
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}
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​
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String res = "";
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​
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for (String word : d) {
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if (word.length() > s.length()) {
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continue;
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}
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​
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if (isSubsequence(s, word)) {
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if (word.length() > res.length() || (word.length() == res.length() && word.compareTo(res) < 0)) {
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res = word;
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}
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}
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}
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return res;
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}
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​
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public boolean isSubsequence(String s, String t) {
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int tIndex = 0;
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for (int i = 0; i < s.length() && tIndex < t.length(); i++) {
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if (s.charAt(i) == t.charAt(tIndex)) {
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++tIndex;
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}
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}
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return tIndex == t.length();
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}
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}
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3. Time & Space Complexity

Two Pointers: 时间复杂度O(n), 空间复杂度O(1)
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Contents
524. Longest Word in Dictionary through Deleting
1. Question
2. Implementation
3. Time & Space Complexity