Jump Game II

       Given an array of non-negative integers, you are initially positioned at the firstindex 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 stepfrom index 0 to 1, then 3 steps to the last index.)

 

       思路：本题的解题思路就是：看第n步最远能走多远。其实本质上类似于广度优先搜索算法，根据nums[0]的值，看第一步最远能走多远，比如到达最远的索引为i，则从1到i之间都属于第一步的范围。再根据nums[1]到nums[i]的值，看第二步能走多远，以此类推。一旦某一步所能到达的最远距离大于等于数组长度，则就得到了结果。代码如下（下面的代码未考虑数组元素值为0的情况）：

#define max(x,y)  (((x)>(y))?(x):(y))
 
int jump(int*nums, int numsSize)
{
    int i;
    int lastmaxnum;
    int curmaxnum = 0;
    int res = 0;
   
    lastmaxnum = 0 + nums[0];
 
    for(i = 1; i <numsSize; i++)
    {
        if(lastmaxnum >= numsSize-1)return res+1;
        if(i < lastmaxnum)
        {
            curmaxnum = max(curmaxnum,i+nums[i]);
        }
        else if(i == lastmaxnum)
        {
            curmaxnum = max(curmaxnum,i+nums[i]);
            res++; 
            lastmaxnum = curmaxnum;        
        }
    }
    return res;
}