Jump Game | & | |

Jump Game |

Given an array of non-negative integers, you is initially positioned at the first index of the array.

Each element of the array represents your maximum jump length is at that position.

Determine if you is able to reach the last index.

Example

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

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

Analysis:

Use a variable to save the currently remaining "steps to jump", if it is less than or equal to 0, it means that you cannot jump further down.

1  Public classSolution {2     /**3 * @param a:a list of integers4 * @return: The Boolean answer5      */6      PublicBoolean Canjump (int[] nums) {7         if(Nums = =NULL)return false;8         intStepsremaining =1;9         Ten          for(inti =0; i < nums.length; i++) { Onestepsremaining = Math.max (Stepsremaining-1, Nums[i]); A             if(Stepsremaining <=0&& I! = nums.length-1)return false; -         } -         return true; the     } -}

Jump Game II

Given an array of non-negative integers, you is initially positioned at the first index of the array.

Each element of the array represents your maximum jump length is at that position.

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

  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 and then steps to the last 3 index.)

Analysis:

When we are at the first point "2", we need to traverse all reachable points, i.e., 3, 1, and then find out at which point (3 or 1) the number of hops left is the highest when we reach the 3rd point.

1  Public classSolution {2     /**3 * @param a:a List of lists of integers4 * @return: An integer5      */6      Public intJumpint[] A) {7         if(A = =NULL|| A.length = =0)return 0;8 9         intPosition =0, Totaljumps =0;Ten         intRemainingjumps =0; One         intjumps = a[0] +1;//assume we jumps in the array at the beginning A  -          while(Position <a.length) { -totaljumps++; the             //find Max remaining jumps in the ' Jump Range ' -              for(inti =1; I <= jumps; i++) { -Remainingjumps = Math.max (A[position], remainingjumps-1); -position++; +                 if(Position >= a.length) Break; -             } +jumps =Remainingjumps; A         } at         returnTotaljumps; -     }  -}

Reference Please specify Source: cnblogs.com/beiyeqingteng/

Jump Game | & | |