题目:

分析:

代码一(暴力解题法,时间复杂度O(n^2)):
public int[] twoSum(int[] numbers, int target) {
int i = 0;
while (i<=numbers.length-2){
int j = 0;
while (i+j+1<=numbers.length-1){
if (numbers[i]+numbers[i+j+1]==target){
return new int[]{i+1, i+j+2};
}
j++;
}
i++;
}
return null;
}
代码二(双指针):
public int[] twoSum1(int[] numbers, int target){
if (numbers.length == 0) return null;
int i = 0;
int j = numbers.length-1;
while (i<j){
int sum = numbers[i]+numbers[j];
if (sum>target){
j--;
}else if (sum<target){
i++;
}else {
return new int[]{i+1, j+1};
}
}
return null;
}