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55. Jump Game && 45. Jump Game II

时间:2015-06-25 10:12:29      阅读:281      评论:0      收藏:0      [点我收藏+]

Jump Game

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.

贪心算法。每次尽可能向前跳,如果最后位置大于长度,则可以跳出。

public class Solution {
  public boolean canJump(int[] nums) {
    int maxIndex = 0;
    for (int i = 0; i < nums.length; i++) {
      if (i > maxIndex) {
        return false;
      }
      maxIndex = Math.max(maxIndex, i + nums[i]);
      }
    return true;
  }
}

 

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.)

依然贪心,每次走到最大的步数即可。

public class Solution {
  public int jump(int[] nums) {
    int jump = 0;
    int i = 0;
    int maxIndex = 0;
    while(maxIndex < nums.length - 1) {
      int curMax = maxIndex;
      for (; i <= curMax; i++) {
        maxIndex = Math.max(maxIndex, nums[i] + i);
      }
      jump++;
      if (curMax == maxIndex) return -1;
    }
    return jump;
  }
}

55. Jump Game && 45. Jump Game II

原文:http://www.cnblogs.com/shini/p/4599186.html

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