164. Maximum Gap
难度:Hard
刷题内容
原题连接
- https://leetcode.com/problems/maximum-gap/
内容描述
Given an unsorted array, find the maximum difference between the successive elements in its sorted form.
Return 0 if the array contains less than 2 elements.
Example 1:
Input: [3,6,9,1]
Output: 3
Explanation: The sorted form of the array is [1,3,6,9], either
(3,6) or (6,9) has the maximum difference 3.
Example 2:
Input: [10]
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.
Note:
You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
Try to solve it in linear time/space.
˼·1 **- ʱ�临�Ӷ�: O(NlgN)*- �ռ临�Ӷ�: O(1)***
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class Solution {
public:
int maximumGap(vector<int>& nums) {
if(nums.size() < 2)
return 0;
sort(nums.begin(),nums.end());
int ans = INT_MIN;
for(int i = 1;i < nums.size();++i)
ans = max(ans,nums[i] - nums[i - 1]);
return ans;
}
};
˼·2 **- ʱ�临�Ӷ�: O(N)*- �ռ临�Ӷ�: O(N)***
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class Solution {
public:
int maximumGap(vector<int>& nums) {
if(nums.size() < 2)
return 0;
int res[nums.size()];
int t[nums.size()][10];
for(int i = 0;i < nums.size();++i)
{
res[i] = i;
for(int j = 0;j < 10;++j)
t[i][j] = 0;
int temp = nums[i],j = 9;
while(temp)
{
t[i][j--] = temp % 10;
temp /= 10;
}
}
for(int i = 9;i >= 0;--i)
{
int bucket[nums.size()],count1[10];
memset(count1,0,sizeof(count1));
for(int j = 0;j < nums.size();++j)
count1[t[res[j]][i]]++;
for(int j = 1;j < 10;++j)
count1[j] += count1[j - 1];
for(int j = nums.size() - 1;j >= 0;--j)
bucket[--count1[t[res[j]][i]]] = res[j];
for(int j = 0;j < nums.size();++j)
res[j] = bucket[j];
}
int ans INT_MIN;
for(int i = 1;i < nums.size();++i)
ans = max(ans,nums[res[i]] - nums[res[i - 1]]);
return ans;
}
};
