尋找和爲定值的兩數(輸出所有對數)

方法一:用HashMap< Integer, Integer> map,一次遍歷,判斷map中是否存在x-a[i],存在則直接返回 a[i] x-a[i],不然將a[i]加入map。java

方法二:排序後,從首尾出發,判斷其和code

public static void main(String[] args) {

		int a[] = { 1, 2, 2, 4,4, 5, 7, 11, 15 };
		useHashmap(a, 15);
		System.out.println("------------");
		useTwoPointer(a, 15);
	}

	private static void useHashmap(int a[], int x) {
		HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
		
		for (int i = 0; i < a.length; i++) {
			
			if( !map.containsKey( a[i])) //添加
				map.put( a[i], 1);
			else map.put( a[i], map.get(a[i])+1);
			
			if (map.containsKey(x - a[i]))  //符合條件
				if( map.get( x-a[i])>1) {
					for( int j=map.get( x-a[i]); j>0; j--)
						System.out.println( a[i] + "," + (x-a[i]));
					map.put( x-a[i], 1);
				}
				else System.out.println( a[i] + "," + (x-a[i]));
		}
		
		
		
	}

	
	private static void useTwoPointer(int a[], int x) {
		Arrays.sort(a);
		int i = 0, j = a.length - 1;

		while (i < j) {
			if (a[i] + a[j] == x) {
				System.out.println(a[i] + "," + a[j]);
				if (a[i + 1] + a[j] == x && i + 1 < j)
					i++;
				else if (a[i] + a[j - 1] == x && i < j - 1)
					j--;
				else
					i++;
			}

			else if (a[i] + a[j] < x)
				i++;
			else
				j--;
		}
	}

結果截圖:排序

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