Jump Game

Description

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.

Notice
This problem have two method which is Greedy and Dynamic Programming.
The time complexity of Greedy method is O(n).
The time complexity of Dynamic Programming method is O(n^2).
We manually set the small data set to allow you pass the test in both ways. 
This is just to let you learn how to use this problem in dynamic programming ways. If you finish it in dynamic programming ways, you can try greedy method to make it accept again.

Example
A = [2,3,1,1,4], return true.
A = [3,2,1,0,4], return false.

Link

Lintcode_ladder

Method

1. x
2. x

Example

1. 1

class Solution {
public:
    /**
     * @param A: A list of integers
     * @return: The boolean answer
     */
    bool canJump(vector<int> nums) {
        // write you code here
        if (nums.empty()) {
            return false;
        }
        // start from the fist steps, we can move
        int curstep = nums[0];
        for (int i = 1; i < nums.size(); ++i) {
            // every loop, we use one step
            --curstep;
            // if we can not move forward and no longer reach the end
            if (curstep < 0) {
                return false;
            }
            // if the current nums[] val can move further, we update the steps
            if (nums[i] > curstep) {
                curstep = nums[i];
            }
        }
        return true;
    }
};

Similar problems

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