根据下述二分查找法，在排好序的数组 1, 3, 6, 9, 17, 31, 39, 52, 61, 79 中查找数值 31，循环 while (left <= right) 执行的次数为（ ）。

int binary_search(vector<int>& nums, int target) {
    int left = 0;
    int right = nums.size() - 1;

    while (left <= right) {
        int mid = left + (right - left) / 2;

        if (nums[mid] == target) {
            return mid;
        }
        else if (nums[mid] < target) {
            left = mid + 1;
        }
        else {
            right = mid - 1;
        }
    }

    return -1; // 如果找不到目标元素，返回-1
}

{{ select(1) }}

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