题目描述

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.

题目思路

• 思路一、递归实现，可惜测试用例很不给力。　超时
  LeetCode 测试用例 74 / 75 test cases passed.

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

• 思路二、贪心算法

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

总结展望