[leetcode] 2055蜡烛之间的盘子 | 前缀和

题目链接

在这里做出处理，让下标从1开始，询问的区间两端也+1
我们开两个数组： n e a r L nearL nearL以及 n e a r R nearR nearR
n e a r L [ i ] nearL[i] nearL[i]表示：在 i i i位置离着最近的左边的   ∣ \ |  ∣符号是在哪一个下标
n e a r R [ i ] nearR[i] nearR[i]表示：在 i i i位置离着最近的右边的   ∣ \ |  ∣符号是在哪一个下标
如果说 s [ i ] = = ′ ∣ ′ s[i] == '|' s[i]==′∣′ 那么说 n e a r L [ i ] = i nearL[i] = i nearL[i]=i并且 n e a r R [ i ] = i nearR[i] = i nearR[i]=i
让左区间等于离着当前位置最近的右边的 |
让右区间等于离着当前位置最近的左边的 |
然后就可以直接开始前缀和了

Code:

class Solution {
public:
    vector<int> platesBetweenCandles(string s, vector<vector<int>>& queries) {
        const int N = s.size();
        int cnt2[N+7];
        vector<int> ans;
        s = "#" + s;
        int tot = queries.size();
        for(vector<int> &v:queries) {
            v[0] += 1;
            v[1] += 1;
        }
        int nearL[N+7], nearR[N+7];
        memset(nearL,0,sizeof nearL);
        memset(nearR,0,sizeof nearR);
        cnt2[0] = 0; 
        for (int i = 1;i <= N;i ++) {
            if(s[i] == '|') {
                cnt2[i] = cnt2[i-1];
                nearL[i] = i;
            } else {
                cnt2[i] = cnt2[i-1] + 1;
                nearL[i] = nearL[i-1];
            }
        }
        for(int i=N;i>=1;i--) {
            if(s[i] == '|') nearR[i] = i;
            else nearR[i] = nearR[i+1];
        }
        for(vector<int> v : queries) {
            int l = v[0], r = v[1];
            if(l == r) {
                ans.push_back(0);
                continue;
            }
            int tl,tr;
            tl = nearR[l];
            tr = nearL[r];
            if(tl >= tr) ans.push_back(0);
            else ans.push_back(cnt2[tr] - cnt2[tl-1]);
        }
        return ans;
    }
};