[Oracle] LeetCode 1326 Minimum Number of Taps to Open to Water a Garden

There is a one-dimensional garden on the x-axis. The garden starts at the point 0 and ends at the point n. (i.e The length of the garden is n).

There are n + 1 taps located at points [0, 1, ..., n] in the garden.

Given an integer n and an integer array ranges of length n + 1 where ranges[i] (0-indexed) means the \(i\)-th tap can water the area [i - ranges[i], i + ranges[i]] if it was open.

Return the minimum number of taps that should be open to water the whole garden, If the garden cannot be watered return -1.

Solution

利用贪心的思想，我们先将区间 \([i-r[i], i+r[i]]\) \(push\_back\) 进去，然后进行 \(sort\). 主要思想就是我们固定左端点的时候，不断地去扩大右端点，然后以此时的右端点为左端点，再次求该时的最远右端点。

点击查看代码

class Solution {
private:
    int ans=0;
    vector<vector<int>> itv;
public:
    int minTaps(int n, vector<int>& ranges) {
        for(int i=0;i<n+1;i++){
            itv.push_back({i-ranges[i], i+ranges[i]});
        }
        sort(itv.begin(), itv.end());
        int st=0, ed=0;
        int i=0;
        while(st<n){
            while(i<itv.size() && itv[i][0]<=st){
                ed = max(ed, itv[i][1]); i++;
            }
            if(st==ed)return -1;
            st = ed;
            ans++;
        }
        return ans;
    }
};