You are given a?0-indexed?binary string?s?and two integers?minJump?and?maxJump. In the beginning, you are standing at index?0, which is equal to?'0'. You can move from index?i?to index?j?if the following conditions are fulfilled:

• i + minJump <= j <= min(i + maxJump, s.length - 1), and

• s[j] == '0'.

Return?true?if you can reach index?s.length - 1?in?s, or?false?otherwise.

?

Example 1:

Input: s = "011010", minJump = 2, maxJump = 3

Output: trueExplanation:In the first step, move from index 0 to index 3. In the second step, move from index 3 to index 5.

Example 2:

Input: s = "01101110", minJump = 2, maxJump = 3

Output: false

?

Constraints:

• 2 <= s.length <= 105

• s[i]?is either?'0'?or?'1'.

• s[0] == '0'

• 1 <= minJump <= maxJump < s.length

• 能過就不奢求了，用dp，只要在i-maxJump 到i-minJump中有一個是true，那么就一定能跳到i，所以dp[i]就一定是true，這里要break一下，不然又要TLE了。。。

Runtime:?2804 ms, faster than?5.06%?of?Java?online submissions for?Jump Game VII.

Memory Usage:?43 MB, less than?85.63%?of?Java?online submissions for?Jump Game VII.