javascript简单算法练习

题目来源于这位大佬的博客：『前端积累』算法汇总【持续更新】 - 风雨后见彩虹 - 博客园 (cnblogs.com)

// 求数组内两数和，跟题目要求不太一样
var nums = [1,2,2,4,3,6,8,5];
var target = 6;
var result = [];
nums.forEach((item, index) => {
  const idx = nums.findIndex(e => ((target - item) === e));
  if (idx !== -1 && idx !== index && idx >= index) {
    result.push({
      start: index,
      end: idx
    })
  }
})
console.log(result);

// 求字符串出现次数最多的字母
const str = 'sfndhvdgndjkfvcvcfwfsdndjfbdhfvdgh';
let max = 0;
let maxKey = '';
newStr = str.split('').reduce((item, next) => {
  item[next] ? item[next] ++ : item[next] = 1;
  if (item[next] > max) {
    max = item[next];
    maxKey = next;
  }
  return item;
}, {})
console.log(newStr);
console.log(max, maxKey);

// 求n位数所有水仙花数
const flower = (num, n) => {
  let test = num;
  let sum = 0;
  for (let i = 1; i <= n; i++) {
    sum += Math.pow(test % 10, n);
    test = (test - test % 10) / 10;
  }
  return sum === num;
}
const allFlower = n => {
  const result = [];
  const min = Math.pow(10, n-1);
  const max = Math.pow(10, n) - 1;
  for (let i = min; i <= max; i++) {
    if (flower(i, n)) {
      result.push(i);
    }
  }
  return result;
}
console.log(allFlower(8)); //8位数需要将近一分钟才能算完，性能问题较差

// 同名去重，不区分大小写
let nameArr = [
  'tao',
  'Tao',
  'Mom',
  'David',
  'david',
  'Meon',
  'mimi'
];
const delSameName = arr => {
  const names = arr.reduce((item, next) => {
    next = next.toLowerCase();
    typeof item.find(e => (next === e)) === 'undefined' && item.push(next);
    return item;
  }, []);
  console.log(names);
}
delSameName(nameArr);

// 反转数字
let num = -50.2;
num = parseFloat(`${num}`.split('').reverse().join('')) * Math.sign(num); // Math.sign()判断正负数，太菜了，这个不懂，参考的大佬代码
console.log(num);

// 求第n个斐波那契数 直接求非数组版
const fbnq = n => {
  if (n === 1) {
    return 0;
  } else if (n === 2) {
    return 1;
  }
  let per = 0;
  let cur = 1;
  let res = 0;
  for (let i = 3; i <= n; i++) {
    res = per + cur;

    // 交换两数
    per += cur;
    cur = per - cur;
    per -= cur;

    // 当前数需相加
    cur = cur + per;
  }
  return res;
}
console.log(fbnq(6));

// 求第n个斐波那契数 数组版
const fbnq = n => {
  let fb = [0, 1];
  if (n === 1) {
    return 0;
  } else if (n === 2) {
    return 1;
  }
  for (let i = 2; i <= n; i++) {
    fb.push(fb[i-2] + fb[i-1]);
  }
  return fb[n-1];
}
console.log(fbnq(6));

// 数组合并排序
let numArr1 = [4,93,23,6,8];
let numArr2 = [3,23,53,1,34];
const newNumArr = numArr1.concat(numArr2).sort((a, b) => a-b);
console.log(newNumArr);

// 求分解质因数
const zysNum = num => {
  const res = [];
  for (let i = 2; i <= num; i++) {
    if (num % i === 0) {
      num = num / i;
      res.push(i);
      i--;
    }
  }
  return res;
}
console.log(zysNum(600));

// 可以删除一个字符的回文 aba和abca都是true
const isHW = hw => {
  let isOnly = 1;
  const newHw = hw.split('');
  const fhw = hw.split('').reverse();
  for (let index = 0; index < (newHw.length / 2); index++) {
    if (newHw[index] !== fhw[index]) { // 判断相反位置的两个是否相等
      if (isOnly) { // 是第一次不相同，删除反转字串该字符后置0
        fhw.splice(index, 1);
        isOnly = 0;
        if (newHw[index] !== fhw[index]) {
          return false;
        }
      } else { // 不是第一次不相同，直接返回false
        return false;
      }
    }
  }
  return true;
}
console.log(isHW('abca'));