leetcode-每日一题2022.3.8 蜡烛之间的盘子

题目

力扣

思路一 二分+前缀和

记录每个蜡烛的位置之前满足条件的*的个数，再对于每一个查询，通过二分，找到左边界右边最近的蜡烛，右边界左边最近的蜡烛，将其中的盘子数量相减。

代码一

class Solution {
public:
    vector<int> idx, sum;
    vector<int> platesBetweenCandles(string s, vector<vector<int>>& queries) {
        vector<int> ans;
        bool flag = false;  //前面是否有蜡烛
        int cnt = 0;
        for(int i = 0; i < s.size(); i++){
            if(s[i] == '|'){
                idx.push_back(i);
                if(!flag){
                    flag = true;
                    sum.push_back(0);
                }else{
                    sum.push_back(cnt);
                }
            }else{
                if(flag) cnt++;
            }
        }
        for(auto query : queries){
            int left = binarySearch(query[0],true), right = binarySearch(query[1],false);
            // cout<<left<<' '<<right<<endl;
            if(right > left)
                ans.push_back(sum[right] - sum[left]);
            else
                ans.push_back(0);
        }
        return ans;
    }
    int binarySearch(int x, bool flag){
        int l = 0, r = idx.size() - 1;
        while(l < r){
            int mid = l + r >> 1;
            if(flag){
                //找到第一个满足idx[i]>=x的i
                if(idx[mid] >= x) r = mid;
                else l = mid + 1;
            }else{
                //找到最后一个满足idx[i]<=x的i
                if(idx[mid + 1] > x) r = mid;
                else l = mid + 1;
            }
        }
        return l;
    }
};

思路二 前缀和

在处理前缀和时，记录每个位置左右最近的蜡烛，避免每次二分。

代码二

class Solution {
public:
    vector<int> platesBetweenCandles(string s, vector<vector<int>>& queries) {
        int n = s.size(), m = queries.size();
        vector<int> l(n), r(n), ans(m), sum(n + 1);
        for(int i = 0, j = n - 1, p = -1, q = -1; i < n; i++, j--){
            if(s[i] == '|') p = i;
            if(s[j] == '|') q = j;
            l[i] = p;
            r[j] = q;
            sum[i + 1] = sum[i] + (s[i] == '*');
        }
        for(int i = 0; i < m; i++){
            int a = queries[i][0], b = queries[i][1];
            int c = r[a], d = l[b];
            if(c < d && c != -1) ans[i] = sum[d + 1] - sum[c + 1];
        }
        return ans;
    }
};