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TwoSumIIInputArrayIsSorted167.java
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TwoSumIIInputArrayIsSorted167.java
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/**
* Given an array of integers that is already sorted in ascending order, find
* two numbers such that they add up to a specific target number.
*
* The function twoSum should return indices of the two numbers such that they
* add up to the target, where index1 must be less than index2.
*
* Note:
* Your returned answers (both index1 and index2) are not zero-based.
* You may assume that each input would have exactly one solution and you may
* not use the same element twice.
*
* Example:
* Input: numbers = [2,7,11,15], target = 9
* Output: [1,2]
* Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
*/
public class TwoSumIIInputArrayIsSorted167 {
public int[] twoSum(int[] numbers, int target) {
if (numbers == null || numbers.length < 2) return new int[2];
int len = numbers.length;
for (int i=0; i<=len-2; i++) {
int idx = binarySearch(numbers, target-numbers[i], i+1, len-1);
if (idx != -1) {
return new int[]{i+1, idx+1};
}
}
return new int[2];
}
private int binarySearch(int[] numbers, int target, int l, int r) {
if (l > r) return -1;
if (l == r) return numbers[l] == target ? l : -1;
int mid = (l + r) / 2;
if (numbers[mid] == target) return mid;
if (numbers[mid] > target) {
return binarySearch(numbers, target, l, mid-1);
} else {
return binarySearch(numbers, target, mid+1, r);
}
}
// using Arrays.binarySearch
public int[] twoSum2(int[] numbers, int target) {
if (numbers == null || numbers.length < 2) return new int[2];
int len = numbers.length;
for (int i=0; i<=len-2; i++) {
int idx = Arrays.binarySearch(numbers, i+1, len, target - numbers[i]);
if (idx >= 0) {
return new int[]{i+1, idx+1};
}
}
return new int[2];
}
}