(Greedy) leetcode 55. 45. Jump Game I II

The meaning of problems: From the array index 0 is, stored in the array is the number of steps can jump at the current index, as arr [0] = 2, 0 represents the index from the index 1 and 2 can reach. Finally, ask if you can reach the final index.

Ideas: The reach: recording a position farthest from the current position that can be reached (counting from 0) takes the maximum value can reach every arrival.

class Solution {
 public :
     BOOL canJump (Vector < int > & the nums) {
         IF (nums.size () < 2 )
             return  to true ;
         int REACH = 0 ;   // REACH: recording from the current position to reach the farthest position 
        for ( int I = 0 ; I <= REACH && I <nums.size (); ++ I) { 
            REACH = max (REACH, I + the nums [I]);
             IF (REACH> = nums.size () - . 1 ) // to reach the tail 
                return  to true ; 
        }
        return false;
    }
};

Meaning of the questions: Returns the minimum number of steps to reach the final position.

 

class Solution {
public:
    int jump(vector<int>& nums) {
        if(nums.empty() || nums.size() <=1)
            return 0;
        //cur_max : 当前位置所能到达的最远位置，next_max: 下一步所能到达的最远位置
        int cur_max = 0, next_max = 0, step = 0, index = 0;
        while(index <= cur_max){
            
            while(index <= cur_max){
                //计算每一步中所能达到的最大位置
                next_max = max(next_max, index+nums[index]);
                index ++;
            }
            cur_max = next_max;
            step ++;
            if(cur_max >= nums.size()-1)
                return step;
        }
        return 0;
    }
};