S019-在既定的数据源生成随机数组

手写-怎么在制定数据源里面生成一个长度为 n 的不重复随机数组能有几种方法时间复杂度多少

第一版

时间复杂度为 O(n^2)

function getTenNum(testArray, n) {
  let result = [];
  for (let i = 0; i < n; ++i) {
    const random = Math.floor(Math.random() * testArray.length);
    const cur = testArray[random];
    if (result.includes(cur)) {
      i--;
      break;
    }
    result.push(cur);
  }
  return result;
}
const testArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14];
const resArr = getTenNum(testArray, 10);

第二版-标记法 / 自定义属性法

时间复杂度为 O(n)

function getTenNum(testArray, n) {
  let hash = {};
  let result = [];
  let ranNum = n;
  while (ranNum > 0) {
    const ran = Math.floor(Math.random() * testArray.length);
    let con = testArray[ran];
    if (!hash[con]) {
      hash[ran] = true;
      result.push(con);
      ranNum--;
    }
  }
  return result;
}
const testArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14];
const resArr = getTenNum(testArray, 10);

第三版-交换法

时间复杂度为 O(n)

function getTenNum(testArray, n) {
  const cloneArr = [...testArray];
  let result = [];
  for (let i = 0; i < n; i++) {
    const ran = Math.floor(Math.random() * (cloneArr.length - i));
    result.push(cloneArr[ran]);
    cloneArr[ran] = cloneArr[cloneArr.length - i - 1];
  }
  return result;
}
const testArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14];
const resArr = getTenNum(testArray, 14);

值得一提的是操作数组的时候使用交换法这种思路在算法里面很常见

最终版

边遍历边删除时间复杂度为 O(n)

function getTenNum(testArray, n) {
  const cloneArr = [...testArray];
  let result = [];
  for (let i = 0; i < n; ++i) {
    const random = Math.floor(Math.random() * cloneArr.length);
    const cur = cloneArr[random];
    result.push(cur);
    cloneArr.splice(random, 1);
  }
  return result;
}
const testArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14];
const resArr = getTenNum(testArray, 14);