[LeetCode] 55. Jump Game 跳跃游戏

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原文链接：http://www.cnblogs.com/lightwindy/p/9640745.html

文章标签：

#数据结构与算法 #java #python

本文深入解析了LeetCode上的跳跃游戏问题，提供了多种解决方案，包括动态规划、贪婪算法和回溯等，通过具体示例展示了如何判断能否从数组起始位置到达末尾。

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;
    }
};

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