codility Frog-River-One

最新推荐文章于 2019-08-17 20:17:15 发布

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本文介绍了一个经典的编程问题——青蛙如何利用连续落在河面上的叶子在规定时间内到达对岸。通过分析给定数组A中叶子落下的位置和时间，实现一个算法来确定青蛙最早何时能够跨越河流。

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A small frog wants to get to the other side of a river. The frog is currently located at position 0, and wants to get to position X. Leaves fall from a tree onto the surface of the river.

You are given a non-empty zero-indexed array A consisting of N integers representing the falling leaves. A[K] represents the position where one leaf falls at time K, measured in minutes.

The goal is to find the earliest time when the frog can jump to the other side of the river. The frog can cross only when leaves appear at every position across the river from 1 to X.

For example, you are given integer X = 5 and array A such that:

  A[0] = 1
  A[1] = 3
  A[2] = 1
  A[3] = 4
  A[4] = 2
  A[5] = 3
  A[6] = 5
  A[7] = 4

In minute 6, a leaf falls into position 5. This is the earliest time when leaves appear in every position across the river.

Write a function:

class Solution { public int solution(int X, int[] A); }

my solution:

class Solution {
public int solution(int X, int[] A) {
// write your code in Java SE 8
int size= A.length;
if (X>size){
return -1;
}

Boolean[] have=new Boolean[X];
for(int i=0;i<X; i++){
have[i] = false;
}

int count =0;

for (int i=0; i< size; i++)
{
if (A[i]<=X && have[A[i]-1]==false){
have[A[i]-1]=true;
count++;
if (count==X)
return i;
}
}

return -1;
}
}