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. Please note that 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.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
一刷
题解： 固定一个值，然后再它的右半部binary search.
Time complexity O(nlogn)
public class Solution {
public int[] twoSum(int[] A, int target) {
int hi = A.length-1;
int[] res = new int[2];
res[0] = -1;
res[1] = -1;
for(int i=0; i<A.length; i++){
int index2 = binarySearch(A, target-A[i], i+1, hi);
if(index2!=-1){
res[0] = i+1;
res[1] = index2+1;
return res;
}
}
return res;
}
private int binarySearch(int[] A, int target, int lo, int hi){
while(lo<=hi){
int mid = lo + (hi - lo)/2;
if(target == A[mid]) return mid;
else if(target>A[mid]) lo = mid+1;
else hi = mid-1;
}
return -1;
}
}
