1、题目
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.
2、wepon的解法
2.1 解析
2.2 代码
/*
法一：桶排序，线性时间空间复杂度
*/
class Solution {
public:
int maximumGap(vector<int> &num) {
if (num.size() < 2) return 0;
//遍历一遍，找出最大最小值
int maxNum = num[0];
int minNum = num[0];
for (int i : num) {
maxNum=max(maxNum,i);
minNum=min(minNum,i);
}
// 每个桶的长度len,向上取整所以加+
int len = (maxNum - minNum) / num.size() + 1;
//桶的个数:(maxNum - minNum) / len + 1,每个桶里面存储属于该桶的最大值和最小值即可，注意这里的最大最小值是局部的
vector<vector<int>> buckets((maxNum - minNum) / len + 1);
for (int x : num) {
int i = (x - minNum) / len;
if (buckets[i].empty()) {
buckets[i].reserve(2);
buckets[i].push_back(x);
buckets[i].push_back(x);
} else {
if (x < buckets[i][0]) buckets[i][0] = x;
if (x > buckets[i][1]) buckets[i][1] = x;
}
}
//gap的计算，For each non-empty buckets p, find the next non-empty buckets q, return min（ q.min - p.max ）
int gap = 0;
int prev = 0;
for (int i = 1; i < buckets.size(); i++) {
if (buckets[i].empty()) continue;
gap = max(gap, buckets[i][0] - buckets[prev][1]);
prev = i;
}
return gap;
}
};
/*
法二：复杂度O（nlogn）
*/
class Solution {
public:
int maximumGap(vector<int> &num) {
if(num.size()<2) return 0;
sort(num.begin(),num.end()); //O(nlogn)
int gap=-1;
for(int i=1;i<num.size();i++){
gap=max(gap,num[i]-num[i-1]);
}
return gap;
}
};