链接:https://www.nowcoder.com/questionTerminal/70610bf967994b22bb1c26f9ae901fa2?f=discussion
来源:牛客网
题目描述:
统计一个数字在排序数组中出现的次数。
思路一:
二分查找,然后向前向后找
代码:
时间复杂度O(logn),空间复杂度O(1)
public class Solution { public int GetNumberOfK(int [] array , int k) { if(array == null || array.length == 0) return 0; int count = 0; int low = 0, high = array.length - 1, mid = 0; while(low <= high){ mid = low + (high -low) / 2; if(array[mid] == k) break; else if(array[mid] > k) high = mid - 1; else low = mid + 1; } if(low <= high){ for(int i = mid;i >= 0;i--){ if(array[i] == k) count++; else break; } for(int i = mid + 1;i < array.length;i++){ if(array[i] == k) count++; else break; } } return count; } }
Runtime:19ms
思路二:(推荐)
看到有序,就应该想到二分查找。找到该数字在数组中第一次、最后一次出现的位置
代码:
public class Solution { public int GetNumberOfK(int [] array , int k) { if(array == null || array.length == 0) return 0; int first = getFirstK(array,k,0,array.length - 1); int last = getLastK(array,k,0,array.length - 1); if(first == -1 || last == -1) return 0; else return last - first + 1; } private int getFirstK(int [] array, int k, int low, int high){ int mid = 0; while(low <= high){ mid = low + (high -low) / 2; if(array[mid] == k){ if(mid > 0 && array[mid - 1] != k || mid == 0) return mid; else high = mid - 1; } else if(array[mid] > k) high = mid - 1; else low = mid + 1; } return -1; } private int getLastK(int [] array, int k, int low, int high){ int mid = 0; while(low <= high){ mid = low + (high -low) / 2; if(array[mid] == k){ if(mid < array.length - 1 && array[mid + 1] != k || mid == array.length - 1) return mid; else low = mid + 1; } else if(array[mid] > k) high = mid - 1; else low = mid + 1; } return -1; } }
Runtime:18ms
思路三:
因为data中都是整数,所以可以稍微变一下,不是搜索k的两个位置,而是搜索(k-0.5)和(k+0.5)这两个数应该插入的位置,然后相减即可。
需要对二分法的代码进行简单的变形修改。
代码:
public class Solution { public int GetNumberOfK(int [] array , int k) { if(array == null || array.length == 0) return 0; return biSearch(array, k + 0.5) - biSearch(array, k - 0.5); } private int biSearch(int [] array, double k){ int low = 0, high = array.length - 1; while(low <= high){ int mid = low + (high -low) / 2; if(array[mid] > k) high = mid - 1; else low = mid + 1; } return low; } }
Runtime:16ms