Given an unsorted array, find the maximum difference between the successive elements in its sorted form.
Try to solve it in linear time/space.
Return 0 if the array contains less than 2 elements.
You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
class Solution {
public:
int maximumGap(vector<int>& nums) {
if(nums.size() < 2) return 0;
int maxn = INT_MIN, minn = INT_MAX;
for(int i = 0; i < nums.size(); i++){
if(nums[i] < minn)
minn = nums[i];
if(nums[i] > maxn)
maxn = nums[i];
}
int gap = maxn - minn+1;
vector<int>vec(gap,0);
for(int i = 0; i < nums.size(); i++){
vec[nums[i] - minn]++;
}
int last = 0, ma = -1;
for(int i = 1; i < gap; i++){
if(vec[i] > 0){
ma = max(ma, i-last);
last = i;
}
}
return ma;
}
};
解題思路:
基數排序(radix sort)/桶排序(bucket sort)
官方版(桶排序):
假設有N個元素A到B。
那麼最大差值不會小於ceiling[(B - A) / (N - 1)]
令bucket(桶)的大小len = ceiling[(B - A) / (N - 1)],則最多會有(B - A) / len + 1個桶
對於數組中的任意整數K,很容易通過算式loc = (K - A) / len找出其桶的位置,然後維護每一個桶的最大值和最小值
由於同一個桶內的元素之間的差值至多爲len - 1,因此最終答案不會從同一個桶中選擇。
對於每一個非空的桶p,找出下一個非空的桶q,則q.min - p.max可能就是備選答案。返回所有這些可能值中的最大值。
class Solution {
public:
int maximumGap(vector<int>& nums) {
if(nums.size() < 2) return 0;
double maxn = INT_MIN, minn = INT_MAX, len = nums.size();
for(int i = 0; i < nums.size(); i++){
if(nums[i] < minn)
minn = nums[i];
if(nums[i] > maxn)
maxn = nums[i];
}
int bucket_size = ceil((maxn - minn)/(len-1)); //桶的大小 出錯2
if(bucket_size == 0) return 0; //出錯1
int bucket_num = (maxn-minn)/bucket_size+1; //桶的個數
vector<vector<int>>vec(bucket_num,vector<int>());
for(int i = 0; i < nums.size(); i++){
vec[(nums[i] - minn)/bucket_size].push_back(nums[i]);
}
int pre_max = 0, res = 0;
for(int i = 0; i < bucket_num; i++){
int cur_min = INT_MAX, cur_max = 0;
for(int j = 0; j < vec[i].size(); j++){
if(vec[i][j] > cur_max)
cur_max = vec[i][j];
if(vec[i][j] < cur_min)
cur_min = vec[i][j];
}
if(vec[i].size() > 0){ //出錯3
if(i > 0) //出錯4
res = max(res, cur_min - pre_max);
pre_max = cur_max;
}
}
return res;
}
};
出錯4:因爲我是計算當前的最小值與前面最大值的差值,因此對於第0號的bucket_num, 它前面的最大值是不知道的,因此需要跳過去,因此要加入if(i>0)