Question: Add Digits
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//Given a non-negative integer num, repeatedly add all its digits until the result has only one digit. | |
//For example: | |
//Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it. | |
//Follow up: | |
//Could you do it without any loop/recursion in O(1) runtime? | |
class AddDigits { | |
public int addDigits(int num) { | |
while(num >= 10) { | |
int temp = 0; | |
while(num > 0) { | |
temp += num % 10; | |
num /= 10; | |
} | |
num = temp; | |
} | |
return num; | |
} | |
} | |
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//Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. | |
//You may assume that the array is non-empty and the majority element always exist in the array. | |
class MajorityElement { | |
public int majorityElement(int[] nums) { | |
if(nums.length == 1) { | |
return nums[0]; | |
} | |
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); | |
for(int current: nums) { | |
if(map.containsKey(current) && map.get(current) + 1 > nums.length / 2) { | |
return current; | |
} else if(map.containsKey(current)) { | |
map.put(current, map.get(current) + 1); | |
} else { | |
map.put(current, 1); | |
} | |
} | |
//no majority element exists | |
return -1; | |
} | |
} |
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