力扣 第 383 场周赛 解题报告 | KMP

链接

前言

一个人能走的多远不在于他在顺境时能走的多快，而在于他在逆境时多久能找到曾经的自己。

T1 修改矩阵

思路：模拟
时间复杂度：$O(mn)$

class Solution:def modifiedMatrix(self, matrix: List[List[int]]) -> List[List[int]]:n, m = len(matrix), len(matrix[0])for j in range(m):mx = -inffor i in range(n):mx = max(mx, matrix[i][j])for i in range(n):if matrix[i][j] == -1:matrix[i][j] = mxans = matrixreturn ans

T2 T4 匹配模式数组的子数组数目

思路： KMP 匹配字符串

class Solution {int ne[1000010];
public:int countMatchingSubarrays(vector<int>& nums, vector<int>& pattern) {int n = nums.size(), m = pattern.size();ne[0] = -1;for (int i = 1, j = -1; i < m; i ++) {while (j != -1 and pattern[i] != pattern[j + 1])    j = ne[j];if (pattern[i] == pattern[j + 1])   j ++;ne[i] = j;}vector<int> v(n - 1);for (int i = 1; i < n; i ++) {if (nums[i] == nums[i-1])   v[i-1] = 0;else if (nums[i] > nums[i-1])   v[i-1] = 1;else    v[i-1] = -1;}int cnt = 0; for (int i = 0, j = -1; i < n - 1; i ++) {while (j != -1 and pattern[j+1] != v[i])    j = ne[j];if (pattern[j+1] == v[i])   j ++;if (j == m - 1) {cnt ++;j = ne[j];}}return cnt;}
};

T3 回文字符串的最大数量
思路： 贪心

class Solution {
public:int maxPalindromesAfterOperations(vector<string>& words) {int odd = 0, even = 0, cnt = 0;unordered_map<char, int> mp;vector<int>  v;for (auto &s: words) {           int n = s.size();v.emplace_back(n);for (char &c: s)    mp[c] ++;}sort(v.begin(), v.end());for (auto it: mp) {int x = it.second;if (x % 2)  {odd ++;even += x - 1;}else {even += x;}}for (auto x: v) {if (x % 2) {if (odd) {odd --;}else {even -= 2;odd ++;}x -= 1; }if (even < x)   break;else {even -= x;cnt ++;}}return cnt;}
}  ;