代码拉取完成，页面将自动刷新
同步操作将从 doocs/leetcode 强制同步，此操作会覆盖自 Fork 仓库以来所做的任何修改，且无法恢复！！！
确定后同步将在后台操作，完成时将刷新页面，请耐心等待。
Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.
Each letter in the magazine string can only be used once in your ransom note.
Example 1:
Input: ransomNote = "a", magazine = "b" Output: false
Example 2:
Input: ransomNote = "aa", magazine = "ab" Output: false
Example 3:
Input: ransomNote = "aa", magazine = "aab" Output: true
Constraints:
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
chars = {}
for i in magazine:
chars[i] = chars.get(i, 0) + 1
for i in ransomNote:
if not chars.get(i):
return False
chars[i] -= 1
return True
class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
int[] chars = new int[26];
for (int i = 0, n = magazine.length(); i < n; ++i) {
int idx = magazine.charAt(i) - 'a';
++chars[idx];
}
for (int i = 0, n = ransomNote.length(); i < n; ++i) {
int idx = ransomNote.charAt(i) - 'a';
if (chars[idx] == 0) return false;
--chars[idx];
}
return true;
}
}
此处可能存在不合适展示的内容，页面不予展示。您可通过相关编辑功能自查并修改。
如您确认内容无涉及 不当用语 / 纯广告导流 / 暴力 / 低俗色情 / 侵权 / 盗版 / 虚假 / 无价值内容或违法国家有关法律法规的内容，可点击提交进行申诉，我们将尽快为您处理。
马建仓 AI 助手
