14. First Position of Target 【easy】
For a given sorted array (ascending order) and a targetnumber, find the first index of this number in O(log n) time complexity.
If the target number does not exist in the array, return -1.
Example
If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.
Challenge
If the count of numbers is bigger than 2^32, can your code work properly?
解法一：
1 class Solution { 2 public: 3 /** 4 * @param nums: The integer array. 5 * @param target: Target number to find. 6 * @return: The first position of target. Position starts from 0. 7 */ 8 int binarySearch(vector<int> &array, int target) { 9 if (array.size() == 0) { 10 return -1; 11 } 12 13 int start = 0; 14 int end = array.size() - 1; 15 16 while (start + 1 < end) { 17 int mid = start + (end - start) / 2; 18 19 if (array[mid] == target) { 20 end = mid; 21 } 22 else if (array[mid] < target) { 23 start = mid; 24 } 25 else if (array[mid] > target) { 26 end = mid; 27 } 28 } 29 30 if (array[start] == target) { 31 return start; 32 } 33 34 if (array[end] = target) { 35 return end; 36 } 37 38 return -1; 39 } 40 };
解法二：
1 class Solution { 2 public: 3 4 int find(vector<int> &array, int start, int end, int target) { 5 if (start > end) { 6 return -1; 7 } 8 9 int mid = start + (end - start) / 2; 10 11 if (array[mid] == target) { 12 13 if (array[mid - 1] != target) { 14 return mid; 15 } 16 17 return find(array, start, mid - 1, target); 18 } 19 else if (array[mid] > target) { 20 return find(array, start, mid - 1, target); 21 } 22 else if (array[mid] < target) { 23 return find(array, mid + 1, end, target); 24 } 25 26 } 27 28 /** 29 * @param nums: The integer array. 30 * @param target: Target number to find. 31 * @return: The first position of target. Position starts from 0. 32 */ 33 int binarySearch(vector<int> &array, int target) { 34 // write your code here 35 36 int start = 0; 37 int end = array.size(); 38 39 return find(array, start, end, target); 40 41 } 42 };