标签：
Jump Game II
问题：
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.
思路：
数组的层次遍历方法
我的代码：
public class Solution { public int jump(int[] nums) { if(nums==null || nums.length<=1) return 0; int start = 0; int end = start+nums[start]; int level = 1; while(end < nums.length-1) { int begin = start+1; int tmpEnd = begin; while(begin <= end) { tmpEnd = Math.max(tmpEnd, nums[begin]+begin); begin++; } if(tmpEnd <= end) return 0; start = end; end = tmpEnd; level++; } return level; } }
他人代码：
public class Solution { public int jump(int[] A) { int[] steps = new int[A.length]; steps[0] = 0; for (int i = 1; i < A.length; i++) { steps[i] = Integer.MAX_VALUE; for (int j = 0; j < i; j++) { if (steps[j] != Integer.MAX_VALUE && j + A[j] >= i) { steps[i] = steps[j] + 1; break; } } } return steps[A.length - 1]; } } // version 2: Greedy public class Solution { public int jump(int[] A) { if (A == null || A.length == 0) { return -1; } int start = 0, end = 0, jumps = 0; while (end < A.length - 1) { jumps++; int farthest = end; for (int i = start; i <= end; i++) { if (A[i] + i > farthest) { farthest = A[i] + i; } } start = end + 1; end = farthest; } return jumps; } }
学习之处：
标签：
原文地址：http://www.cnblogs.com/sunshisonghit/p/4513684.html
