RansomNote

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 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  wil…

ransom-note

https://leetcode.com/problems/ransom-note/ public class Solution { public boolean canConstruct(String ransomNote, String magazine) { char[] chRan = ransomNote.toCharArray(); char[] chMag = magazine.toCharArray(); Arrays.sort(chRan); Arrays.sort(chMag…

[LeetCode] Ransom Note 赎金条

 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  wil…

Leetcode分类刷题答案&心得

Array 448.找出数组中所有消失的数要求:整型数组取值为 1 ≤ a[i] ≤ n,n是数组大小,一些元素重复出现,找出[1,n]中没出现的数,实现时时间复杂度为O(n),并不占额外空间思路1:(discuss)用数组下标标记未出现的数,如出现4就把a[3]的数变成负数,当查找时判断a的正负就能获取下标 tips:注意数组溢出 public List<Integer> findDisappearedNumbers(int[] nums) { List<Integer> d…

LeetCode之383. Ransom Note

-------------------------------------------- 思路就是进行频率统计. 统计一下第二个字符串字符出现次数++统计一下第一个字符串中字符出现次数--如果出现负数说明第二个中的字符不够用的. AC代码如下: public class Solution { public boolean canConstruct(String ransomNote, String magazine) { int book[]=new int[26]; for(int i=0;i…

leetcode 383. Ransom Note

 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  wil…

leetcode bugfree note

463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…

LeetCode: Ransom Note

public class Solution { public boolean canConstruct(String ransomNote, String magazine) { int[] ransomNum = new int[256]; int[] magNum = new int[256]; for (int i = 0; i < 256; i++) { ransomNum[i] = magNum[i] = 0; } for (int i = 0; i < ransomNote.len…

String Start！

(1)Ransom Note 解题思路: 题目叫做Ransom Note,勒索信.勒索信,为了不暴露字迹,就从杂志上搜索各个需要的字母,组成单词来表达的意思.这样来说,题目也就清晰了,判断杂志上的字是否能够组成勒索信需要的那些字符. 这里需要注意的就是杂志上的字符只能被使用一次,不过不用考虑大小写的问题. 有一种最简单的理解就是对于ransomNote里每个字符出现的次数必须小于或者等于该字符在magazine出现的次数. 原理就是列出了magazine的字母表,然后算出了出现个数,然后遍历ra…

LeetCode:Ransom Note_383

LeetCode:Ransom Note [问题再现] 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  magaz…