|
| 1 | +/** |
| 2 | + * Given an array of non-negative integers, |
| 3 | + * you are initially positioned at the first index of the array. |
| 4 | + * Each element in the array represents your maximum jump length at that position. |
| 5 | + * Your goal is to reach the last index in the minimum number of jumps. |
| 6 | + * For example: |
| 7 | + * Given array A = [2,3,1,1,4] |
| 8 | + * The minimum number of jumps to reach the last index is 2. |
| 9 | + * (Jump 1 step from index 0 to 1, then 3 steps to the last index.) |
| 10 | + */ |
| 11 | +public class JumpGame { |
| 12 | + public static void main(String[] args) { |
| 13 | + int[] a1 = new int[]{2, 3, 1, 1, 4}; |
| 14 | + int[] a2 = new int[]{2, 3, 1, 1, 0,4}; |
| 15 | + |
| 16 | + jumpGame(a1); |
| 17 | + jumpGame(a2); |
| 18 | + |
| 19 | + |
| 20 | + } |
| 21 | + |
| 22 | + public static void jumpGame(int[] a) { |
| 23 | + if (a.length == 0) return; |
| 24 | + |
| 25 | + int[] preStep = new int[a.length]; |
| 26 | + |
| 27 | + for (int i = 0; i < preStep.length; i++) |
| 28 | + preStep[i] = -1; |
| 29 | + |
| 30 | + boolean reachEnd = false; |
| 31 | + int furthest = a[0]; |
| 32 | + int i = 0; |
| 33 | + while (i <= furthest) { |
| 34 | + furthest = Math.max(furthest, a[i] + i); |
| 35 | + for (int j = 0; j <= furthest; j++) { |
| 36 | + if (j < a.length) { |
| 37 | + if (preStep[j] == -1) |
| 38 | + preStep[j] = i; |
| 39 | + else { |
| 40 | + if (preStep[j] > i) |
| 41 | + preStep[j] = i; |
| 42 | + } |
| 43 | + } |
| 44 | + } |
| 45 | + if (furthest >= a.length - 1) { |
| 46 | + reachEnd = true; |
| 47 | + break; |
| 48 | + } |
| 49 | + i++; |
| 50 | + } |
| 51 | + |
| 52 | + if (!reachEnd) { |
| 53 | + System.out.println("unreachable!"); |
| 54 | + } else { |
| 55 | + int end = a.length - 1; |
| 56 | + while (end != 0) { |
| 57 | + System.out.print(a[end] +"->"); |
| 58 | + end = preStep[end]; |
| 59 | + } |
| 60 | + System.out.print(a[end]); |
| 61 | + System.out.println(); |
| 62 | + } |
| 63 | + |
| 64 | + } |
| 65 | +} |
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