找LeetCode上简单的题目来撑过30天啦(DAY16)

ok，今天挑战同一张证照第三次失败了，有够难过的，但至少分述有越来越接近啦，再接再厉罗

题号：55 标题：Jump Game 难度：Medium

You are given an integer array nums. You are initially positioned at the array's first index, and each element in the array represents your maximum jump length at that position.
Return true if you can reach the last index, or false otherwise.

Example 1:
Input: nums = [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:
Input: nums = [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible to reach the last index.

Constraints:
• 1 <= nums.length <= 104
• 0 <= nums[i] <= 105

class Solution {
    public boolean canJump(int[] nums) {
        int i=0,j=0,k=0;
        int[] temp = new int[nums.length+1];
        ArrayList<Integer> arrlist = new ArrayList<Integer>();
        
        for(i=0;i<nums.length;i++){
            arrlist.add(nums[i]);
        }
        if(!arrlist.contains(0)){
            return true;
        }
        
        if(nums[0]==0){
            if(nums.length==1){
                return true;
            }else{
                return false;
            }
        }
        int count0=0,chk0=0;
        for(i=nums.length-1;i>=0;i--){
            if(nums[i]==0){
                count0++;
                chk0=1;
               
            }else if(chk0==1 && nums[i]!=0 ){
                k=0;
                for(j=i;j>=0;j--){
                    if(nums[j]>(count0+k)){
                        chk0=0;
                        count0=0;
                        break;
                    }else if(count0+i+1==nums.length && nums[j]==(count0+k)){
                        chk0=0;
                        count0=0;
                        break;
                    }else if(j==0){
                        return false;
                    }
                    k++;
                }
            }
        }
        
        return true;
    }
}

DAY16心得
证照跟铁人赛还有上班三头烧，我已经再用测资测试程序了，看看我把CASE一个个判断出来
1.没有0的一定到的了=>TRUE
2.第一个值为0，一定到不了=>FALSE
3.接下来就是判断连续几个0前面的值要能够跨过去，能够跨过去就继续搜寻有没有0，跨不过去就代表到不了=>FALSE