一:題目描述:
二:Code:
兩種方法:
import java.util.Arrays;
import org.junit.Test;
public class totalArray {
public void printResult(int a, int b) {
System.out.println(a + "+" + b);
}
// 時間複雜度O(n^2)
public void twoSum(int[] array, int s) {
int len = array.length;
for (int i = 0; i <= len - 1; i++) {
for (int j = i + 1; j <= len - 1; j++) {
if (array[i] + array[j] == s) {
printResult(array[i], array[j]);
break;
}
}
}
}
//// 時間複雜度爲線性
public void twoSum02(int[] array, int s) {
int i = 0;
int j = array.length - 1;
while (i < j) {
int sum = array[i] + array[j];
if (sum == s) {
printResult(array[i], array[j]);
i++;
j--;
} else if (sum > s) {
j--;
} else {
i++;
}
}
}
// 給定的數組已經排好序了,所以用二分查找,時間複雜度爲O( nlog(n) )
public void twoSum03(int[] array, int s) {
int n = array.length;
for (int i = 0; i < n - 1; i++) {
int another = s - array[i];
if (Arrays.binarySearch(array, i + 1, n, another) >= i + 1) {
printResult(array[i], another);
}
}
}
@Test
public void testTwosum() {
int[] array = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
int s = 10;
twoSum03(array, s);
}
}