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.
Using Greedy Algorithm

bool canJump(int A[], int n) {

if (0 == n)

            return false;

        int cur_max = A[0];

        int i = 0;

        while (i <= cur_max) {

            cur_max = max(cur_max, i + A[i]);

            

            if(cur_max >= n - 1)

                return true;

                

            ++i;

        }

        return false;

    }

bool canJump(int A[], int n) {       

        if (1 == n)

            return true;

        int i = 0;

        while (i < n)

        {

            int currMax = A[i] + i;

            

            if (0 == A[i])

                return false;

            if (currMax >= n - 1)

                return true;

                

            int tmpMax = 0;

            for (int j = i + 1; j <= currMax; ++j)

            {

                if (A[j] + j > tmpMax)

                {

                    tmpMax = A[j] + j;

                    i = j;

                }

            }

        }

        

        return (i >= n - 1);

    }

DP
bool canJump(int A[], int n) {
        bool ATag[n];
        for (int i = n - 1; i >= 0; --i)
        {
            ATag[i] = false;
            int maxReach = A[i] + i;
           
            if (maxReach >= n - 1)
            {
                ATag[i] = true;
                continue;
            }
           
            for (int j = maxReach; j > i; --j)
            {
                if (ATag[j])
                {
                    ATag[i] = true;
                    break;
                }
            }
        }
        return ATag[0];
    }
Read full article from Jump Game

Jump Game II

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.
Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]
The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

Using Greedy Algorithm: O(n)

int jump(int A[], int n) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        

        if (n == 1)

            return 0;

        

        int jumps = 0;

        int currMax = 0;

        int i = 0;

        

        while (i < n)

        {

            currMax = A[i] + i;

            if (A[i] > 0)

                ++jumps;

            else

                return 0;

                

            if (currMax >= n - 1)

                return jumps;

            

            int tmpMax = 0;

            for (int j = i + 1; j <= currMax; ++j)

            {

                if (A[j] + j >= tmpMax)

                {

                    tmpMax = A[j] + j;

                    i = j;

                }

            }

        }

        

        return jumps;

    }

Code from http://blog.csdn.net/linhuanmars/article/details/21356187
public int jump(int[] A) {
    if(A==null || A.length==0)
        return 0;
    int lastReach = 0;
    int reach = 0;
    int step = 0;
    for(int i=0;i<=reach&&i<A.length;i++)
    {
        if(i>lastReach)
        {
            step++;
            lastReach = reach;
        }
        reach = Math.max(reach,A[i]+i);
    }
    if(reach<A.length-1)
        return 0;
    return step;
}

Using DP, Code from Jump Game II

int jump(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(n <= 1)
            return 0;
        const int noWay = n + 1;
        int *jumps = new int[n];
        jumps[n-1] = 0;
        for(int i = n - 2; i >= 0; -- i) {
            int lenJump = A[i];
            int minJumps = noWay;
            for(int j = i + 1; j <= i + lenJump && j < n; ++ j) {
                if(jumps[j] + 1 < minJumps)
                    minJumps = jumps[j] + 1;
            }
            jumps[i] = minJumps;
        }
        return jumps[0];
    }

Also refer to http://www.darrensunny.me/leetcode-jump-game-ii/
Read full article from Jump Game II