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LeetCode:Jump Game

发布时间:2016-12-3 21:49:36 编辑:www.fx114.net 分享查询网我要评论
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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. Determine if you are able to reach the last index. For example: A = [2,3,1,1,4], return true. A = [3,2,1,0,4], return false. 脑子不好使,起初我写了个O(N*N)复杂度的算法超时了。后来更正了算法。 算法思想是从前往后遍历,不断更新能跳到的最远距离。算法O(N)时间复杂度。 package leetcode; import java.io.IOException; import java.io.InputStreamReader; import java.io.StreamTokenizer; public class JumpGame { static { System.setIn(JumpGame.class.getResourceAsStream("/data/leetcode/JumpGame.txt")); } private static StreamTokenizer stdin = new StreamTokenizer(new InputStreamReader(System.in)); private static int readInt() throws IOException { stdin.nextToken(); return (int) stdin.nval; } public static void main(String[] args) throws IOException { JumpGame s = new JumpGame(); int[] a = new int[25003]; for (int i = 0; i < 25003; i++) { a[i] = readInt(); } long start = System.currentTimeMillis(); System.out.println(s.canJump(a)); System.out.println(System.currentTimeMillis() - start); } public boolean canJump(int[] A) { if (A.length <= 1) { return true; } int maxLen = 0; int index = 0; while (index <= maxLen) { int a = A[index]; if (index + a > maxLen) { maxLen = index + a; } if (maxLen >= A.length - 1) { return true; } index++; } return false; } }

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