PAT甲级 The Dominant Color （技巧+java版map）

The Dominant Color 

链接：https://www.nowcoder.com/questionTerminal/0495013675774f008541ea371eb5af17
来源：牛客网

Behind the scenes in the computer's memory, color is always talked about as a series of 24 bits of information for each pixel. In an image, the color with the largest proportional area is called the dominant color. A strictly dominant color takes more than half of the total area. Now given an image of resolution M by N (for example, 800x600), you are supposed to point out the strictly dominant color.

输入描述:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: M (<=800) and N (<=600) which are the resolutions of the image. Then N lines follow, each contains M digital colors in the range [0, 2²⁴). It is guaranteed that the strictly dominant color exists for each input image. All the numbers in a line are separated by a space.

输出描述:

For each test case, simply print the dominant color in a line.

示例1

输入

5 3
0 0 255 16777215 24
24 24 0 0 24
24 0 24 24 24

输出

24
题意：
输出给出的数中的众数
题解：技巧版
　　 因为答案数的个数一定会超过总数的一半 所以可以当当前数是答案数的时候，计数器加一，不是的时候，计数器减一，最后不能完全抵消的就是答案。

import java.util.Scanner;
public class Main {
    static int n,m;
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        m = cin.nextInt();
        n = cin.nextInt();
        int ans = 0,cnt=0;
        for(int i=0;i<m;i++) {
            for(int j=0;j<n;j++) {
                int x = cin.nextInt();
                if(cnt == 0) {
                    ans = x;
                    cnt = 1;
                }
                else {
                    if(x == ans) {
                        cnt++;
                    }
                    else {
                        cnt--;
                    }
                }
            }
        }
        System.out.println(ans);
    }
}

然后是java的map版

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;
 
 
public class Main{
    static int n,m;
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        m = cin.nextInt();
        n = cin.nextInt();
        int ans = 0,cnt=0;
        int value = 0;int x;
        Map<Integer, Integer> map = new HashMap<Integer,Integer>();
        for(int i=0;i<m;i++) {
            for(int j=0;j<n;j++) {
                 x = cin.nextInt();
                if(map.get(x) == null) {
                    value = 1;
                }
                else {
                    value = (int)map.get(x);
                    value++;
                }
                map.put(x, value);
            }
        }
        Iterator<Map.Entry<Integer, Integer>> it = map.entrySet().iterator();
        int max = 0,res=0;
        while(it.hasNext()) {
            Entry<Integer, Integer> entry = it.next();
            if(entry.getValue()>max) {
                res = entry.getKey();
                max = entry.getValue();
            }
        }
        System.out.println(res);
    }
}