雙指針+了一點小小小的優化:
class Solution {
public int[] twoSum(int[] numbers, int target) {
int left=0;
int right=numbers.length-1;
while(left<=right){
int mediate=left+(right-left)/2;
if(numbers[left]+numbers[right]==target) return new int[] {left+1,right+1};
else if(numbers[left]+numbers[right]<target){
if(numbers[mediate]+numbers[right]<target) left=mediate+1;
else if(numbers[mediate]+numbers[right]>target) left=left+1;
else return new int[] {mediate+1,right+1};
}
else if(numbers[left]+numbers[right]>target){
if(numbers[mediate]+numbers[left]>target) right=mediate-1;
else if(numbers[mediate]+numbers[left]<target) right=right-1;
else return new int[] {left+1,mediate+1};
}
}
return null;
}
}
看見了一個很妙的寫法,可以用二分法確定右邊的範圍,然後用雙指針來確定最終結果。
class Solution {
public int[] twoSum(int[] numbers, int target) {
if(numbers==null) return null;
int tail=Dichotomy(numbers,target-numbers[0]);
int head=0;
while(head<tail){
int sum=numbers[head]+numbers[tail];
if(sum==target) return new int[]{head+1,tail+1};
else if(sum<target) head++;
else tail--;
}
return null;
}
private int Dichotomy(int[] numbers,int target){//找到了最接近目標值的下標
int left=0;
int right=numbers.length-1;
while(left<=right){
int mediate=left+(right-left)/2;
if(numbers[mediate]==target) return mediate;
else if(numbers[mediate]>target) right=mediate-1;
else left=mediate+1;
}
return left-1;//即返回的num[left-1]<target
}
}
如果nums[i]>target-nums[0]
的話,則nums[i]以及右邊肯定不是所求。
我們需要nums[i]<=target-nums[0]
。而Dichotomy解決的就是這個問題。
也瞭解了,由於+的優先級高於>>,所以8+(10)>>1
結果是9,而不是13.