leetCode 55.Jump Game(跳躍游戲) 解題思路和方法
Jump Game
Given an array of non-negative integers, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Determine if you are able to reach the last index.
For example:
A = [2,3,1,1,4], return true.
A = [3,2,1,0,4], return false.
思路：這題相比於jump Game II多了0的填充，有可能是無法到達最終點的。代碼運用貪心思想，當無論怎麼走只能走到0的時候返回false。 具體代碼如下：
public class Solution {
public boolean canJump(int[] nums) {
if(nums.length == 0)
return false;
int i = 0;
//判斷有沒有0，沒有0的肯定能達到
while(i < nums.length){
if(nums[i] == 0){
break;
}
i++;
}
//沒有0，肯定能達到
if(i == nums.length){
return true;
}
i = 0;
while(i < nums.length){
if(i + nums[i] >= nums.length - 1)
return true;
if(nums[i] == 0)
return false;
int max = 0;
int index = 0;
//下一步能前進最大的步驟
for(int j = i+1; j - i <= nums[i]; j++){
if(max < j - i + nums[j]){
max = j - i + nums[j];
index = j;
}
}//走到下一步的索引
i = index;
}
return true;
}
}
