LintCode 462 [Total Occurrence of Target]

原题

给出一个目标数字和一个升序整数数组，返回目标数字在数组中出现的次数。

• 给出 [1, 3, 3, 4, 5] 并且 target = 3, 返回 2.
• 给出 [2, 2, 3, 4, 6] 并且 target = 4, 返回 1
• 给出 [1, 2, 3, 4, 5] 并且 target = 6, 返回 0

解题思路

• 二分法， 分两步：找到第一个出现的位置， 找到最后一个出现的位置。就可以计算出出现频率

完整代码

class Solution:
    # @param {int[]} A an integer array sorted in ascending order
    # @param {int} target an integer
    # @return {int} an integer
    def totalOccurrence(self, A, target):
        # Write your code here
        if not A:
            return 0
            
        start, end = 0, len(A) - 1
        while start + 1 < end:
            mid = start + (end - start) / 2
            if A[mid] <= target:
                start = mid
            else:
                end = mid
        
        if A[end] == target:
            right = end
        elif A[start] == target:
            right = start
        else:
            return 0
        
        start, end = 0, len(A) - 1
        while start + 1 < end:
            mid = start + (end - start) / 2
            if A[mid] >= target:
                end = mid
            else:
                start = mid
        
        if A[start] == target:
            left = start
        elif A[end] == target:
            left = end
        else:
            return 0
            
        return right - left + 1