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]
思路：這題很類似Jump Game II; 把能夠jump的距離，轉換成arr, 距離全部存在left開始的array裏面（不是中心），問題就轉換爲 Jump Game II; 代碼一模一樣；
class Solution {
public int minTaps(int n, int[] ranges) {
int[] arr = new int[n + 1];
for(int i = 0; i < ranges.length; i++) {
if(ranges[i] == 0) continue;
int left = Math.max(0, i - ranges[i]);
arr[left] = Math.max(arr[left], i + ranges[i]);
}
int end = 0, farCanReach = 0, step = 0;
for(int i = 0; i < arr.length && end < n; end = farCanReach) {
step++;
while(i < arr.length && i <= end) {
farCanReach = Math.max(farCanReach, arr[i++]);
}
if(end == farCanReach) {
return -1;
}
}
return step;
}
}