leetcode -- 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.

[解题思路]

1.一维DP，jump[i]表示到达位置i剩余的最大步数

public boolean canJump(int[] A) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int len = A.length;
        //using DP
        int[] jump = new int[len];
        jump[0] = 0;
        
        for(int i = 1; i < len; i++){
            jump[i] = Math.max(jump[i - 1], A[i - 1]) - 1;
            if(jump[i] < 0){
                return false;
            }
        }
        return true;
    }

 2.二维DP

f(i)表示是否可以从position 0到达i

f(i) = OR 'f(j) && A[j] + j >= i' j = 0,...,i - 1

时间复杂为O(N^2), 大数据集会超时

public boolean canJump(int[] A) {
        if(A.length == 0){
            return true;
        }
        int len = A.length;
        boolean[] f = new boolean[len];
        f[0] = true;
        for(int i = 1; i < len; i++){
            for(int j = 0; j < i; j++){
                f[i] = f[i] || (f[j] && (A[j] + j >= i));
            }
        }
        return f[len - 1];
    }

 3.Greedy

维护一个maxDis变量，表示当前可以到达的最大距离

当maxDis >= len - 1表示可以到达

每经过一点都将maxDis 与 i + A[i]进行比较，更新最大值

public boolean canJump(int[] A) {
        if(A.length == 0){
            return true;
        }
        int len = A.length;
        int maxDis = 0;
        for(int i = 0; i < len; i++){
            
            if(maxDis < i){
                return false;
            }
            maxDis = Math.max(maxDis, i + A[i]);
            if(maxDis >= len - 1){
                return true;
            }
        }
        
        return false;
    }