[LintCode] Maximum Gap 求最大间距
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.
Notice
You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
Given [1, 9, 2, 5], the sorted form of it is[1, 2, 5, 9], the maximum gap is between 5and 9 = 4.
Sort is easy but will cost O(nlogn) time. Try to solve it in linear time and space.
LeetCode上的原题,请参见我之前的博客Maximum Gap。
class Solution {
public:
/**
* @param nums: a vector of integers
* @return: the maximum difference
*/
int maximumGap(vector<int> nums) {
if (nums.empty()) return ;
int mx = INT_MIN, mn = INT_MAX, n = nums.size();
for (int d : nums) {
mx = max(mx, d);
mn = min(mn, d);
}
int size = (mx - mn) / n + ;
int bucket_num = (mx - mn) / size + ;
vector<int> bucket_min(bucket_num, INT_MAX);
vector<int> bucket_max(bucket_num, INT_MIN);
set<int> s;
for (int d : nums) {
int idx = (d - mn) / size;
bucket_min[idx] = min(bucket_min[idx], d);
bucket_max[idx] = max(bucket_max[idx], d);
s.insert(idx);
}
int pre = , res = ;
for (int i = ; i < n; ++i) {
if (!s.count(i)) continue;
res = max(res, bucket_min[i] - bucket_max[pre]);
pre = i;
}
return res;
}
};
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