新手如何做代理手机优化软件下载

前言：这个题目其实就是考察前缀和，正常情况下开二维数组来记录，但是也可以优化成一位的位运算
我们顺便可以学习一下如何进行查询二进制串中1的个数

class Solution {
public:vector<bool> canMakePaliQueries(string s, vector<vector<int>> &queries) {int n = s.length(), q = queries.size();vector<array<int, 26>> sum(n + 1);for (int i = 0; i < n; i++) {sum[i + 1] = sum[i];sum[i + 1][s[i] - 'a']++;}vector<bool> ans(q);for (int i = 0; i < q; i++) {auto &query = queries[i];int left = query[0], right = query[1], k = query[2], m = 0;for (int j = 0; j < 26; j++)m += (sum[right + 1][j] - sum[left][j]) % 2; // 奇数+1，偶数+0if((right-left+1)&1) m--;ans[i] = m  <= k*2;}return ans;}
};

优化后的代码

class Solution {
public:vector<bool> canMakePaliQueries(string s, vector<vector<int>>& queries) {int n = s.size();vector<int> count(n + 1);for (int i = 0; i < n; i++) {count[i + 1] = count[i] ^ (1 << (s[i] - 'a'));}vector<bool> res;for (auto& query : queries) {int l = query[0], r = query[1], k = query[2];int bits = 0, x = count[r + 1] ^ count[l];while (x > 0) {x &= x - 1;bits++;}res.push_back(bits <= k * 2 + 1);}return res;}
};