[LeetCode] 1189. Maximum Number of Balloons

Given a string text, you want to use the characters of text to form as many instances of the word "balloon" as possible.

You can use each character in text at most once. Return the maximum number of instances that can be formed.

Example 1:

Input: text = "nlaebolko"
Output: 1

Example 2:

Input: text = "loonbalxballpoon"
Output: 2

Example 3:

Input: text = "leetcode"
Output: 0

Constraints:

1 <= text.length <= 104
text consists of lower case English letters only.

“气球” 的最大数量。

给你一个字符串 text，你需要使用 text 中的字母来拼凑尽可能多的单词 "balloon"（气球）。

字符串 text 中的每个字母最多只能被使用一次。请你返回最多可以拼凑出多少个单词 "balloon"。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-number-of-balloons
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

这道题的思路不难，就是用 hashmap 统计。这里我给出两种实现，思路都是统计每个字母的出现次数。

时间O(n)

空间O(n)

Java实现一

 1 class Solution {
 2     public int maxNumberOfBalloons(String text) {
 3         HashMap<Character, Integer> map = new HashMap<>();
 4         for (char c : text.toCharArray()) {
 5             map.put(c, map.getOrDefault(c, 0) + 1);
 6         }
 7         
 8         int count = 0;
 9         while (true) {
10             int countB = map.getOrDefault('b', 0);
11             int countA = map.getOrDefault('a', 0);
12             int countL = map.getOrDefault('l', 0);
13             int countO = map.getOrDefault('o', 0);
14             int countN = map.getOrDefault('n', 0);
15             if (countB >= 1 && countA >= 1 && countL >= 2 && countO >= 2 && countN >= 1) {
16                 count++;
17                 map.put('b', countB - 1);
18                 map.put('a', countA - 1);
19                 map.put('l', countL - 2);
20                 map.put('o', countO - 2);
21                 map.put('n', countN - 1);
22             } else {
23                 break;
24             }
25         }
26         return count;
27     }
28 }

Java实现二

 1 class Solution {
 2     public int maxNumberOfBalloons(String text) {
 3         int[] map = new int[26];
 4         for (char c : text.toCharArray()) {
 5             map[c - 'a']++;
 6         }
 7         int min = map[1];                        // for b
 8         min = Math.min(min, map[0]);            // for a
 9         min = Math.min(min, map[11] / 2);        // for l /2 
10         min = Math.min(min, map[14] / 2);        // similarly for o/2
11         min = Math.min(min, map[13]);            // for n
12         return min;        
13     }
14 }

LeetCode 题目总结