二分查找

details

/**
 *
 * @param {number[]} nums
 * @param {number} target
 * @returns {number}
 */
function binarySearch(nums, target) {
  //
  let left = 0;
  let right = nums.length - 1;
  while (left <= right) {
    let mid = left + Math.ceil((right - left) / 2);
    if (nums[mid] === target) {
      return mid;
    } else if (nums[mid] > target) {
      right = mid - 1;
    } else {
      left = mid + 1;
    }
  }
}