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.
Example 1:
Input: n = 5, ranges = [3,4,1,1,0,0]
Output: 1
Explanation: The tap at point 0 can cover the interval [-3,3]
The tap at point 1 can cover the interval [-3,5]
The tap at point 2 can cover the interval [1,3]
The tap at point 3 can cover the interval [2,4]
The tap at point 4 can cover the interval [4,4]
The tap at point 5 can cover the interval [5,5]
Opening Only the second tap will water the whole garden [0,5]
Example 2:
Input: n = 3, ranges = [0,0,0,0]
Output: -1
Explanation: Even if you activate all the four taps you cannot water the whole garden.
Example 3:
Input: n = 7, ranges = [1,2,1,0,2,1,0,1]
Output: 3
Example 4:
Input: n = 8, ranges = [4,0,0,0,0,0,0,0,4]
Output: 2
Example 5:
Input: n = 8, ranges = [4,0,0,0,4,0,0,0,4]
Output: 1
Constraints:
-
1 <= n <= 10^4 -
ranges.length == n + 1 -
0 <= ranges[i] <= 100
题解：
之前按点算的，一直不对，看讨论区是按区间算的，dp[1]代表0-1这个区间。
动态规划问题：先遍历每个喷头的喷射范围，然后计算范围内需要的喷头最小数。
class Solution {
public:
int minTaps(int n, vector<int>& ranges) {
vector<int> dp(n + 1, 1000);
dp[0] = 0;
for (int i = 0; i <= n; i++) {
int left = max(0, i - ranges[i]);
int right = min(n, i + ranges[i]);
for (int j = left; j <= right; j++) {
dp[j] = min(dp[j], dp[left] + 1);
}
}
if (dp[n] >= 1000) {
return -1;
}
return dp[n];
}
};
