167. Two Sum II - Input array is sorted
原题链接:https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/
这道题目蛮有意思的,有点难度哦:
import java.util.Arrays;
/**
* Created by clearbug on 2018/2/26.
*/
public class Solution {
public static void main(String[] args) {
Solution s = new Solution();
System.out.println(Arrays.toString(s.twoSum1(new int[]{2, 7, 11, 15}, 9)));
System.out.println(Arrays.toString(s.twoSum1(new int[]{2, 7, 11, 15}, 18)));
System.out.println(Arrays.toString(s.twoSum2(new int[]{2, 7, 11, 15}, 9)));
System.out.println(Arrays.toString(s.twoSum2(new int[]{2, 7, 11, 15}, 18)));
}
/**
* 方法一:最黄最暴力
*
* @param numbers
* @param target
* @return
*/
public int[] twoSum1(int[] numbers, int target) {
for (int i = 0; i < numbers.length; i++) {
for (int j = i + 1; j < numbers.length; j++) {
if (numbers[i] + numbers[j] == target) {
return new int[]{i + 1, j + 1};
}
}
}
return null;
}
/**
* 方法二:分析方法一可知,方法一有两层循环,时间复杂度极高。然后,我想可以使用类似二分查找的思想来优化,但是却想不出究竟怎样完美的用上
* 二分查找思想,然后就去看了下提交区运行时间最低的答案,瞬间明白了。。。后来,写了下实现,才发现刚才那瞬间自己根本没明白,现在才
* 稍微明白点了。。。
*
* @param numbers
* @param target
* @return
*/
public int[] twoSum2(int[] numbers, int target) {
int start = 0, end = numbers.length - 1;
while (start < end) {
if (numbers[start] + numbers[end] == target) {
return new int[]{start + 1, end + 1};
} else if (numbers[start] + numbers[end] < target) {
// move start forward
start = smallestLargerOrFirstEqual(numbers, start, end, target - numbers[end]);
} else { // numbers[start] + numbers[end] > target
// move end backward
end = largestSmallerOrLastEqual(numbers, start, end, target - numbers[start]);
}
}
return null;
}
private int smallestLargerOrFirstEqual(int[] numbers, int start, int end, int target) {
while (start <= end) {
int middle = (start + end) / 2;
if (numbers[middle] == target) {
return middle;
} else if (numbers[middle] < target) {
start = middle + 1;
} else {
end = middle - 1;
}
}
return start;
}
private int largestSmallerOrLastEqual(int[] numbers, int start, int end, int target) {
while (start <= end) {
int middle = (start + end) / 2;
if (numbers[middle] == target) {
return middle;
} else if (numbers[middle] < target) {
start = middle + 1;
} else {
end = middle - 1;
}
}
return end;
}
}