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Given an array nums of size n, return the majority element.
The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.
Example 1:
<pre>Input: nums = [3,2,3] Output: 3 </pre>
Example 2:
<pre>Input: nums = [2,2,1,1,1,2,2] Output: 2 </pre>
Constraints:
n == nums.length1 <= n <= 5 * 10<sup>4</sup>-2<sup>31</sup> <= nums[i] <= 2<sup>31</sup> - 1
Follow-up: Could you solve the problem in linear time and in O(1) space?</div>
题目大意：
求数组中的众数
解题思路：
编程之美的水王法
解题步骤：
N/A
注意事项：
Python代码：
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12def majorityElement(self, nums: List[int]) -> int:
candidate, count = 0, 0
for i in range(len(nums)):
if count == 0:
candidate = nums[i]
count += 1
continue
if nums[i] == candidate:
count += 1
else:
count -= 1
return candidate
算法分析：
时间复杂度为O(n)，空间复杂度O(1)
