题目链接：164. 最大间距 - 力扣（LeetCode）
解析：
排序后相邻数字最大间距 要 >= ceil((max - min) / (n - 1))
然后分桶，保持每个桶的max和min即可
class Solution { public:int maximumGap(vector<int>& nums) {if (nums.size() == 1) return 0;int min_v = nums[0], max_v = nums[0];int n = nums.size();int same_flag = true;for (int i = 1; i < n; i++) {min_v = min(nums[i], min_v);max_v = max(nums[i], max_v);if (nums[i] != nums[i - 1]) {same_flag = false;}}if (same_flag) return 0;int segment = ceil((max_v - min_v) / (double) (n - 1));vector<vector<int>> terminal;terminal.reserve(n + 1);for (int i = 0; i <= n; i++) {vector<int> inner;terminal.push_back(inner);terminal[i].push_back(-1);terminal[i].push_back(0x7fffffff);}for (int i = 0; i < n; i++) {int d = (nums[i] - min_v) / segment;terminal[d][0] = max(terminal[d][0], nums[i]);terminal[d][1] = min(terminal[d][1], nums[i]);}int pre_max = -1;int ret = -1;for (int i = 0; i <= n; i++) {if (terminal[i][0] == -1) continue;cout << terminal[i][1] << " " << terminal[i][0] << endl;if (pre_max == -1) {pre_max = terminal[i][0];} else {ret = max(ret, terminal[i][1] - pre_max);pre_max = terminal[i][0];}}return ret;} };