力扣2055.蜡烛之间的盘子

• 预处理每个元素左右最近的蜡烛下标

  • 同时求前缀和
  • 遍历每个询问
  • 找到左右端点对应的内部的最近蜡烛(最大区间)

•   class Solution {
    public:
        vector<int> platesBetweenCandles(string s, vector<vector<int>>& queries) {
            vector<int> l(s.length(), 0), r(s.length(), 0);
            vector<int> sum(s.length() + 1, 0);
            int p = -1,q = -1;
            for(int i=0,j=s.size()-1;i<s.size();i++,j--)
            {
                //pq为最近的蜡烛下标
                if(s[i] == '|') p = i;
                if(s[j] == '|') q = j;
                l[i] = p,r[j] = q;
                sum[i+1] = sum[i] + (s[i] == '*' ? 1 : 0);
            }
    
            vector<int> res(queries.size(),0);
            for(int i=0;i<queries.size();i++)
            {
                int a = queries[i][0] , b = queries[i][1];
                //最大区间的左右端点
                int c = r[a],d = l[b];
                if(c != -1 && c <= d) res[i] = sum[d + 1] - sum[c];
            }
            return res;
        }
    };