189. Rotated arrays, circular substitution, JavaScript, detailed comments

Original link: leetcode-cn.com/problems/ro…

Answer:

This method refers to the official solution and the rotation array transposition, detailed diagrams, and provides detailed notes to help understand.

/ * * *@param {number[]} nums
 * @param {number} k
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var rotate = function (nums, k) {
  k = k % nums.length; // k can be longer than the array length,
  let count = 0; // Count the number of iterations

  /* * this loop implements two effects: * 1. Count is the number of moves currently made, and when count reaches the last digit of the array, the flip is over. * 2. Start is the starting position of the current flip. To handle both cases, refer to the condition */ of the while loop
  for (let start = 0; count < nums.length; start++) {
    let currentMoveIndex = start; // Record the starting position of the current movement
    let currentMoveItem = nums[start]; // Record the start value of the current movement

    do {
      // Calculate the position to move to, cycle to the next k positions of the current position
      let nextMoveIndex = (currentMoveIndex + k) % nums.length;
      // Cache the value of the location to be moved
      let temp = nums[nextMoveIndex];
      // Move the current value to the next location
      nums[nextMoveIndex] = currentMoveItem;
      // The value of temp is the value to be moved next time
      currentMoveItem = temp;
      NextMoveIndex is the starting value of the next move
      currentMoveIndex = nextMoveIndex;
      count++; // Count moves
    } while (
      Nums. length and k's greatest common divisor is not 1, assuming m. * Essentially, after m while loops, start and currentMoveIndex will be equal. * Move start backward to continue the loop. Start and currentMoveIndex will not be equal if nums.length is 1. * This loop will execute until currentMoveIndex=nums.length-1, which means that the array has been traversed exactly once. * /
      start !== currentMoveIndex
    );
  }
};
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