35回文排序

解法1

/**
 * @param {string} s
 * @return {boolean}
 */
var canPermutePalindrome = function(s) {
    let obj = {};
    for(let i of s){
        obj[i] = obj[i] ? 0 : 1;
    }
    return Object.values(obj).filter(it=> it === 1) < 2;
};

解法2

/**
 * @param {string} s
 * @return {boolean}
 */
var canPermutePalindrome = function(s) {
  let obj = {};
  for(let i=0; i<s.length; i++){
    let char = s[i];
    if(obj[char])
      delete obj[char]
    else
      obj[char] = 1;
  }
  return Object.keys(obj).length <= 1;
};