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. (Jump1step from index 0 to 1, then3steps to the last index.)
从后往前，倒推，判断当前位置到达最后所需要的最少steps。PS：也可以使用DP，从前往后递推。public int jump(int[] A) { if (A == null || A.length == 1) { return 0; } int len = A.length; int[] num = new int[len]; num[len - 1] = 0; for (int k = len - 2; k >= 0; k--) { int step = A[k]; int min = Integer.MAX_VALUE - 1; for (int i = k + step; i > k; i--) { if (i >= len - 1) { min = 0; i = k; } else if (num[i] == 1) { min = 1; i = k; } else if (min > num[i]){ min = num[i]; } } num[k] = min + 1; // System.out.println("K :" + k + " "+ num[k]); } return num[0]; }
原文地址：http://blog.csdn.net/u010378705/article/details/30068697
