383.赎金信

发表于 分类于 阅读次数：

给定一个赎金信 (ransom) 字符串和一个杂志(magazine)字符串，判断第一个字符串 ransom 能不能由第二个字符串 magazines 里面的字符构成。如果可以构成，返回 true ；否则返回 false。

(题目说明：为了不暴露赎金信字迹，要从杂志上搜索各个需要的字母，组成单词来表达意思。杂志字符串中的每个字符只能在赎金信字符串中使用一次。)

示例 1：

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输入：ransomNote = "a", magazine = "b"
输出：false

示例  2:

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输入：ransomNote = "aa", magazine = "ab"
输出：false

代码实现: 循环

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/**
 * @param {string} ransomNote
 * @param {string} magazine
 * @return {boolean}
 */
var canConstruct = function (ransomNote, magazine) {
  if (ransomNote.length > magazine.length) return false;
  let obj = {};
  for (let i = 0; i < magazine.length; i++) {
    let z = magazine[i];
    if (obj[z]) {
      obj[z] += 1;
    } else {
      obj[z] = 1;
    }
  }
  for (let i = 0; i < ransomNote.length; i++) {
    let z = ransomNote[i];
    if (obj[z] > 0) {
      obj[z] -= 1;
    } else {
      return false;
    }
  }
  return true;
};

用数组替换 hashmap:

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var canConstruct = function (ransomNote, magazine) {
  let arr = new Array(26).fill(0);
  for (let i = 0; i < magazine.length; i++) {
    arr[magazine.codePointAt(i) - "a".codePointAt(0)]++;
  }
  for (let i = 0; i < ransomNote.length; i++) {
    arr[ransomNote.codePointAt(i) - "a".codePointAt(0)]--;
    if (arr[ransomNote.codePointAt(i) - "a".codePointAt(0)] < 0) {
      return false;
    }
  }
  return true;
};