Majority Number III

Given an array of integers and a number k, the majority number is the number that occurs more than 1/k of the size of the array.

Find it.

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Example
Given [3,1,2,3,2,3,3,4,4,4] and k=3, return 3.

Note
There is only one majority number in the array.

Challenge
O(n) time and O(k) extra space

这道题跟Lintcode: Majority Number II思路很像，那个找大于1/3的，最多有两个candidate，这个一样，找大于1/k的，最多有k-1个candidate

维护k-1个candidate 在map里面，key为数字值，value为出现次数。先找到这k-1个candidate后，扫描所有元素，如果该元素存在在map里面，update map；如果不存在，1： 如果map里面有值为count= 0，那么删除掉这个元素，加入新元素；2：map里面没有0出现，那么就每个元素的count--

注意：有可能map里有多个元素count都变成了0，只用删掉一个就好了。因为还有0存在，所以下一次再需要添加新元素的时候不会执行所有元素count-1, 而是会用新元素替代那个为0的元素

这道题因为要用map的value寻找key，所以还可以用map.entrySet(), return Map.Entry type, 可以调用getKey() 和 getValue()

public class Solution {
    /**
     * @param nums: A list of integers
     * @param k: As described
     * @return: The majority number
     */
    public int majorityNumber(ArrayList<Integer> nums, int k) {
        // write your code
        if (nums==null || nums.size()==0 || k<=0) return Integer.MIN_VALUE;
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        int i = 0;
        for (; i<nums.size(); i++) {
            int key = nums.get(i);
            if (map.containsKey(key)) {
                map.put(key, map.get(key)+1);
            }
            else {
                map.put(key, 1);
                if (map.size() >= k) break;
            }
        }
        while (i < nums.size()) {
            int key = nums.get(i);
            if (map.containsKey(key)) {
                map.put(key, map.get(key)+1);
            }
            else {
                if (map.values().contains(0)) { //map contains value 0
                    map.put(key, 1); // add new element to map
                    //delete key that has value 0
                    int zeroKey = 0;
                    for (int entry : map.keySet()) {
                        if (map.get(entry) == 0) {
                            zeroKey = entry;
                            break;
                        }
                    }
                    map.remove(zeroKey);
                }
                else {
                    for (int nonzeroKey : map.keySet()) {
                        map.put(nonzeroKey, map.get(nonzeroKey)-1);
                    }
                }
            }
            i++;
        }
        
        HashMap<Integer, Integer> newmap = new HashMap<Integer, Integer>();
        int max = 0;
        int major = 0;
        for (int j=0; j<nums.size(); j++) {
            int cur = nums.get(j);
            if (!map.containsKey(cur)) continue;
            if (newmap.containsKey(cur)) {
                newmap.put(cur, newmap.get(cur)+1);
            }
            else {
                newmap.put(cur, 1);
            }
            if (newmap.get(cur) > max) {
                major = cur;
                max = newmap.get(cur);   
            }
        }
        return major;
    }
}

　　

 if (map.values().contains(0))