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  • [LeetCode] 55. Jump Game 跳跃游戏

    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.

    Example 1:

    Input: [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: [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.

    解法1:DP

    解法2:贪婪算法Greedy

    Java:

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

    Python:

    class Solution:
        # @param A, a list of integers
        # @return a boolean
        def canJump(self, A):
            reachable = 0
            for i, length in enumerate(A):
                if i > reachable:
                    break
                reachable = max(reachable, i + length)
            return reachable >= len(A) - 1
    

    C++: Backtrack

    class Solution {
    public:
        bool canJump(int A[], int n) {
            int last=n-1,i,j;
            for(i=n-2;i>=0;i--){
                if(i+A[i]>=last)last=i;
            }
            return last<=0;
        }
    };
    

    C++: DP

    class Solution {
    public:
        bool canJump(vector<int>& nums) {
            vector<int> dp(nums.size(), 0);
            for (int i = 1; i < nums.size(); ++i) {
                dp[i] = max(dp[i - 1], nums[i - 1]) - 1;
                if (dp[i] < 0) return false;
            }
            return dp.back() >= 0;
        }
    };
    

    C++: Greedy

    class Solution {
    public:
        bool canJump(vector<int>& nums) {
            int n = nums.size(), reach = 0;
            for (int i = 0; i < n; ++i) {
                if (i > reach || reach >= n - 1) break;
                reach = max(reach, i + nums[i]);
            }
            return reach >= n - 1;
        }
    };
    

        

    类似题目:

    [LeetCode] 45. Jump Game II 跳跃游戏 II 

    All LeetCode Questions List 题目汇总

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  • 原文地址:https://www.cnblogs.com/lightwindy/p/9640745.html
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