题目
本次主要涉及 3 个题目。前两个题目是 Majority Element I 和 II，第三个是一个类似的应用。
Majority Element
Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
You may assume that the array is non-empty and the majority element always exist in the array.
Example 1:
Input: [3,2,3]
Output: 3
Example 2:
Input: [2,2,1,1,1,2,2]
Output: 2
Majority Element II
Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Note: The algorithm should run in linear time and in O(1) space.
Example 1:
Input: [3,2,3]
Output: [3]
Example 2:
Input: [1,1,1,3,3,2,2,2]
Output: [1,2]
String Compression
Given an array of characters, compress it in-place.
The length after compression must always be smaller than or equal to the original array.
Every element of the array should be a character (not int) of length 1.
After you are done modifying the input array in-place, return the new length of the array.
Follow up: Could you solve it using only O(1) extra space?
Example 1:
Input: [“a”,”a”,”b”,”b”,”c”,”c”,”c”]
Output: Return 6, and the first 6 characters of the input array should be: [“a”,”2”,”b”,”2”,”c”,”3”]
Explanation: “aa” is replaced by “a2”. “bb” is replaced by “b2”. “ccc” is replaced by “c3”.
Example 2:
Input: [“a”]
Output: Return 1, and the first 1 characters of the input array should be: [“a”]
Explanation: Nothing is replaced.
Example 3:
Input: [“a”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”]
Output: Return 4, and the first 4 characters of the input array should be: [“a”,”b”,”1”,”2”].
Explanation: Since the character “a” does not repeat, it is not compressed. “bbbbbbbbbbbb” is replaced by “b12”. Notice each digit has it’s own entry in the array.
Note:
- All characters have an ASCII value in [35, 126].
- 1 <= len(chars) <= 1000.
解法
算法复杂度 O(n)，空间复杂度 O(1)
思路
第二题的思路相对直接会有一点绕。其仍然是 count 的想法，但是维护了两个计数器。
字符压缩题中也使用了类似的 count 思想，但是需要注意在循环之后还需要将最后一个可能的重复字符给补上。
代码
class Solution {
public:
int majorityElement(vector<int>& nums) {
int element = 0;
int count = 0;
for (int& num : nums) {
if (count == 0) {
count = 1;
element = num;
} else if (num == element) {
++count;
} else {
--count;
}
}
return element;
}
};
class Solution {
public:
vector<int> majorityElement(vector<int>& nums) {
int cnt1 = 0;
int cnt2 = 0;
int val1, val2;
for (int& num : nums) {
if (cnt1 != 0 && val1 == num) {
++cnt1;
} else if (cnt2 != 0 && val2 == num) {
++cnt2;
} else if (cnt1 == 0) {
cnt1 = 1;
val1 = num;
} else if (cnt2 == 0) {
cnt2 = 1;
val2 = num;
} else {
--cnt1;--cnt2;
}
}
/* cout << "val1: " << val1 << " cnt1: " << cnt1 << endl; */
/* cout << "val2: " << val2 << " cnt2: " << cnt2 << endl; */
cnt1 = cnt2 = 0;
for (int& num: nums) {
if (num == val1) ++cnt1;
else if (num == val2) ++ cnt2;
}
vector<int> result;
int low = nums.size() / 3;
if (cnt1 > low) result.push_back(val1);
if (cnt2 > low) result.push_back(val2);
return result;
}
};
class Solution {
private:
void put_count(vector<char>& chars, int cur_count, int& cur_pos) {
string s = to_string(cur_count);
int length = s.size();
for (char& c : s)
chars[cur_pos++] = c;
}
public:
int compress(vector<char>& chars) {
int length = chars.size();
if (length == 0) return 0;
int cur_pos = 0;
char cur_char = ' ';
int cur_count = -1;
for (int j = 0; j < length; ++j) {
if (cur_count == -1) {
chars[cur_pos] = chars[j];
cur_count = 1;
cur_char = chars[j];
/* position to store number */
++cur_pos;
} else if (chars[j] == cur_char) {
++cur_count;
} else {
if (cur_count > 1)
put_count(chars, cur_count, cur_pos);
cur_char = chars[j];
chars[cur_pos] = chars[j];
cur_count = 1;
cur_pos++;
}
}
if (cur_count > 1) {
int n = 1;
int m = cur_count;
put_count(chars, cur_count, cur_pos);
}
return cur_pos;
}
};
本文完