LeetCode:Jump Game II

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

.
(Jump

1

step
from index 0 to 1, then

3

steps
to the last index.)

这题很有意思，算法时间复杂度O(N).

package leetcode;

import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.util.ArrayList;
import java.util.List;

public class JumpGameII {

	static {
        System.setIn(JumpGameII.class.getResourceAsStream("/data/leetcode/JumpGameII.txt"));
    }
    private static StreamTokenizer stdin = new StreamTokenizer(new InputStreamReader(System.in));
    private static int readInt() throws IOException {
        stdin.nextToken();
        return (int) stdin.nval;
    }
	/**
	 * @param args
	 */
	public static void main(String[] args) throws IOException {
		JumpGameII s = new JumpGameII();
    	int[] m = new int[5];
    	for (int i = 0; i < m.length; i++) {
    		
    			m[i] = readInt();
    		
    	}
    	long start = System.currentTimeMillis();
		System.out.println(s.jump(m));
		System.out.println("time:" + (System.currentTimeMillis() - start));
	}
	
	public int jump(int[] A) {
		int len = A.length;
		if (len == 0 || len == 1) {
			return 0;
		}
		if (A[0] >= len - 1) {
			return 1;
		}
		
		List<Integer> list = new ArrayList<Integer>();
		list.add(A[0]);
		int index = 1;
		int step = 1;
		while (index < A.length - 1) {
			while (index > list.get(step - 1)) {
				step++;
			}
			int jump = A[index];
			int jumpTo = index + jump;
			if (jumpTo > list.get(step - 1)) {
				if (list.size() == step) {
					list.add(jumpTo);
				} else {
					if (jumpTo > list.get(step)) {
						list.set(step, jumpTo);
					}
				}
				
				if (jumpTo >= len - 1) {
					return step + 1;
				}
			}
			
			index++;
		}
		
		return -1;
	}

}