华为OD机试：攀登者

题目描述

攀登者喜欢寻找各种地图，并且尝试攀登到最高的山峰。

地图表示为一维数组，数组的索引代表水平位置，数组的元素代表相对海拔高度。其中数组元素0代表地面。

例如：[0,1,2,4,3,1,0,0,1,2,3,1,2,1,0]，代表如下图所示的地图，地图中有两个山脉位置分别为 1,2,3,4,5 和 8,9,10,11,12,13，最高峰高度分别为 4,3。最高峰位置分别为3,10。

一个山脉可能有多座山峰(高度大于相邻位置的高度，或在地图边界且高度大于相邻的高度)。

登山时会消耗登山者的体力(整数)，

上山时，消耗相邻高度差两倍的体力

下山时，消耗相邻高度差一倍的体力

平地不消耗体力

登山者体力消耗到零时会有生命危险。

例如，上图所示的山峰：

从索引0，走到索引1，高度差为1，需要消耗 2 * 1 = 2 的体力

从索引2，走到索引3，高度差为2，需要消耗 2 * 2 = 4 的体力

从索引3，走到索引4，高度差为1，需要消耗 1 * 1 = 1 的体力。

攀登者想要评估一张地图内有多少座山峰可以进行攀登，且可以安全返回到地面，且无生命危险。

例如上图中的数组，有3个不同的山峰，登上位置在3的山可以从位置0或者位置6开始，从位置0登到山顶需要消耗体力 1 * 2 + 1 * 2 + 2 * 2 = 8，从山顶返回到地面0需要消耗体力 2 * 1 + 1 * 1 + 1 * 1 = 4 的体力，按照登山路线 0 → 3 → 0 需要消耗体力12。攀登者至少需要12以上的体力（大于12）才能安全返回。

输入描述

第一行输入为地图一维数组

第二行输入为攀登者的体力

输出描述

确保可以安全返回地面，且无生命危险的情况下，地图中有多少山峰可以攀登。

用例

JS算法源码
const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;


void (async function () {
  const heights = (await readline()).split(",").map(Number);
  const strength = parseInt(await readline());
  console.log(getResult(heights, strength));
})();

function getResult(heights, strength) {
  const idxs = new Set();


  climb(heights, strength, idxs, true);

  climb(heights.reverse(), strength, idxs, false);

  return idxs.size;
}

function climb(heights, strength, idxs, direction) {

  let j = 0;
  while (j < heights.length && heights[j] != 0) {
    j++;
  }

  let cost = 0; 
  for (let i = j + 1; i < heights.length; i++) {
    if (heights[i] == 0) {
      cost = 0;
      continue;
    }

    const diff = heights[i] - heights[i - 1]; 

    if (diff > 0) {
      cost += diff * 3;
     
      if (i + 1 >= heights.length || heights[i] > heights[i + 1]) {
        if (cost < strength) {
          if (direction) {
            idxs.add(i);
          } else {
            idxs.add(heights.length - i - 1); 
          }
        }
      }
    } else if (diff < 0) {
      cost -= diff * 3;

    }
  }
}

Java算法源码
import java.util.Arrays;
import java.util.HashSet;
import java.util.Scanner;

public class Main {
  
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);

    int[] heights = Arrays.stream(sc.nextLine().split(",")).mapToInt(Integer::parseInt).toArray();
    int strength = Integer.parseInt(sc.nextLine());

    System.out.println(getResult(heights, strength));
  }

  public static int getResult(int[] heights, int strength) {

    HashSet<Integer> idxs = new HashSet<>();

    climb(heights, strength, idxs, true);

    reverse(heights);
    climb(heights, strength, idxs, false);

    return idxs.size();
  }

  public static void climb(int[] heights, int strength, HashSet<Integer> idxs, boolean direction) {

    int j = 0;
    while (j < heights.length && heights[j] != 0) {
      j++;
    }

    int cost = 0; 
    for (int i = j + 1; i < heights.length; i++) {
      if (heights[i] == 0) {
        cost = 0;
        continue;
      }

      int diff = heights[i] - heights[i - 1]; 

      if (diff > 0) {
        cost += diff * 3;
   
        if (i + 1 >= heights.length || heights[i] > heights[i + 1]) {
          if (cost < strength) {

            if (direction) {
              idxs.add(i);
            } else {
              idxs.add(heights.length - i - 1); 
            }
          }
        }

      } else if (diff < 0) {
        cost -= diff * 3;
 
      }
    }
  }

  public static void reverse(int[] nums) {
    int i = 0;
    int j = nums.length - 1;

    while (i < j) {
      int tmp = nums[i];
      nums[i] = nums[j];
      nums[j] = tmp;
      i++;
      j--;
    }
  }
}

Python算法源码

heights = list(map(int, input().split(",")))
strength = int(input())


def climb(idxs, direction):
    j = 0
    while j < len(heights) and heights[j] != 0:
        j += 1

    cost = 0

    for i in range(j + 1, len(heights)):
        if heights[i] == 0:
            cost = 0
            continue

        diff = heights[i] - heights[i - 1]

        if diff > 0:
            cost += diff * 3

            if i + 1 >= len(heights) or heights[i] > heights[i + 1]:
                if cost < strength:
                    if direction:
                        idxs.add(i)
                    else:
                        idxs.add(len(heights) - i - 1)  

        elif diff < 0:
            cost -= diff * 3
       


def getResult():
    idxs = set()

    climb(idxs, True)

    heights.reverse()
    climb(idxs, False)

    return len(idxs)


print(getResult())

C算法源码
#include <stdio.h>

#define MAX_SIZE 100000

int canClimb[MAX_SIZE] = {0};
int canClimb_count = 0;

void climb(const int heights[], int heights_size, int strength, int direction) {

    int j = 0;
    while (j < heights_size && heights[j] != 0) {
        j++;
    }

    int cost = 0; 
    for (int i = j + 1; i < heights_size; i++) {
        if (heights[i] == 0) {
            cost = 0;
            continue;
        }

        int diff = heights[i] - heights[i - 1]; 

        if (diff > 0) {
            cost += diff * 3;
            if (i + 1 >= heights_size || heights[i] > heights[i + 1]) {
                if (cost < strength) {
                    int idx = direction ? i : heights_size - i - 1;

                    if(!canClimb[idx]) {
                        canClimb[i] = 1;
                        canClimb_count++;
                    }
                }
            }

        } else if (diff < 0) {
            cost -= diff * 3;

        }
    }
}

void reverse(int nums[], int nums_size) {
    int i = 0;
    int j = nums_size - 1;

    while (i < j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;

        i++;
        j--;
    }
}

int getResult(int heights[], int heights_size, int strength) {
    climb(heights, heights_size, strength, 1);

    reverse(heights, heights_size);
    climb(heights, heights_size, strength, 0);

    return canClimb_count;
}

int main() {
    int heights[MAX_SIZE];
    int heights_size = 0;

    while (scanf("%d", &heights[heights_size++])) {
        if (getchar() != ',') break;
    }

    int strength;
    scanf("%d", &strength);

    printf("%d\n", getResult(heights, heights_size, strength));

    return 0;
}