问题:
难度:medium
说明:
给出一个数组 nums,和一个长度 K,要求返回一个长度 K 的数组,该数组属于原数组 nums 的子序列,然后要求返回数组是最具有竞争力,其实就是子序列的数字相对原数组顺序,前端序的都是最小的。
题目连接:https://leetcode.com/problems/find-the-most-competitive-subsequence/
输入范围:
1 <= nums.length <= 105
0 <= nums[i] <= 109
1 <= k <= nums.length
输入案例:
Example 1:
Input: nums = [3,5,2,6], k = 2
Output: [2,6]
Explanation: Among the set of every possible subsequence: {[3,5], [3,2], [3,6], [5,2], [5,6], [2,6]}, [2,6] is the most competitive.
Example 2:
Input: nums = [2,4,3,3,5,4,9,6], k = 4
Output: [2,3,3,4]
我的代码:
因为输入长度不可能全排列的,所以用单调栈,然后需要对比下单调栈长度和对比的起始位置就好。
Java:
class Solution {
public int[] mostCompetitive(int[] nums, int k) {
int len = nums.length, index = 0;
int[] res = new int[k];
Arrays.fill(res, Integer.MAX_VALUE);
for(int i = 0; i < len; i ++) {
boolean flag = true; // 判断是否将数据插入标记
int top = index, begin = len - i >= k ? 0 : k - len + i; // 计算单调栈比较开端索引
while(top - 1 >= begin && res[top - 1] > nums[i]) { // top是比最后一个元素多1,并且要先 -1 判断再进行 -1
res[-- top] = nums[i];
flag = false;
}
if(!flag) index = top + 1; // 重新给index赋值
else if(index < k) res[index ++] = nums[i];
}
return res;
}
}
C++:
class Solution {
public:
vector<int> mostCompetitive(vector<int>& nums, int k) {
vector<int> res(k, INT_MAX); // 填满最大值
int index = 0, len = nums.size();
for(int i = 0; i < len; i ++) {
bool flag = true;
int top = index, begin = len - i >= k ? 0 : k - len + i;
while (top - 1 >= begin && nums[i] < res[top - 1]) {
res[-- top] = nums[i];
flag = false;
}
if(!flag) index = top + 1;
else if(index < k) res[index ++] = nums[i];
}
return res;
}
};