/*** @param {string} ransomNote* @param {string} magazine* @return {boolean}*/varcanConstruct=function(ransomNote, magazine){const strArr =newArray(26).fill(0), base ="a".charCodeAt();for(const s of magazine){// 记录magazine里各个字符出现次数strArr[s.charCodeAt()- base]++;}for(const s of ransomNote){// 对应的字符个数做--操作const index = s.charCodeAt()- base;if(!strArr[index])returnfalse;// 如果没记录过直接返回falsestrArr[index]--;}returntrue;};
方法二：字符统计
/*** @param {string} ransomNote* @param {string} magazine* @return {boolean}*/varcanConstruct=function(ransomNote, magazine){if(ransomNote.length > magazine.length){returnfalse;}const cnt =newArray(26).fill(0);for(const c of magazine){cnt[c.charCodeAt()-'a'.charCodeAt()]++;}for(const c of ransomNote){cnt[c.charCodeAt()-'a'.charCodeAt()]--;if(cnt[c.charCodeAt()-'a'.charCodeAt()]<0){returnfalse;}}returntrue;};
方法三：暴力解法
/*** @param {string} ransomNote* @param {string} magazine* @return {boolean}*/varcanConstruct=function(ransomNote, magazine){for(let i =0; i < magazine.length; i++){for(let j =0; j < ransomNote.length; j++){// 在ransomNote中找到和magazine相同的字符if(magazine[i]=== ransomNote[j]){ransomNote.splice(ransomNote[j],1);// ransomNote删除这个字符串}}}// 如果ransomNote为空，则说明magazine的字符可以组成ransomNoteif(ransomNote.length ===0){returntrue;console.log("magazine的字符可以组成ransomNote");}returnfalse;console.log("magazine的字符不可以组成ransomNote");};let a =canConstruct(["a","a","c"],["a","a","b","c"]);
console.log(a);
