LeetCode|Two Sum-java

题目

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution.

Example：

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,

return [0, 1].

需要注意的：数组并不是排序的数组

public class Solution {
/**
* 思路一：
* 将数组拷贝出一份，对数组进行排序，然后使用head和end指针移动查
* 找答案，最后在原数组中查找相应的相应的位置。
* 空间复杂度：o(n)
* 时间复杂度：o(nlogn+n+n) = o(nlogn)
*
* */
public int[] twoSum(int[] nums, int target) {
int[] temp = Arrays.copyOfRange(nums, 0, nums.length);
Arrays.sort(temp);

int head = 0;
int end = temp.length - 1;
int sum = temp[head] + temp[end];
while (sum != target && head < end) {
sum = temp[head] + temp[end];
if (sum < target) {
head++;
} else if (sum > target) {
end--;
}
}
if (head != end) {
int[] result = {-1, -1};
for (int i = 0; i < nums.length; i++) {
if (nums[i] == temp[head] && result[0] == -1) {
result[0] = i;
} else if (nums[i] == temp[end] && result[1] == -1) {
result[1] = i;
} else if (result[0] != -1 && result[1] != -1) {
break;
}
}
Arrays.sort(result);
return result;
}
return null;
}
}

public class Solution {
/**
* 思路二：
* 将数组的值和相应的位置信息存储到hashMap中，Key存储数组的值，Value存贮该值得位置，
* 然后遍历整个数组，因为数组中有且只有一对数字的合等于目标数字，所以将每次遍历的数字与目标数字相减，
* 查看hasMap中是否有相应的结果
* 空间复杂度：o(n)
* 时间复杂度：o(n);
*
* */
public int[] twoSum(int[] nums, int target) {
if (nums == null || nums.length == 0) return null;
HashMap<Integer, Integer> hashMap = new HashMap<Integer, Integer>();
for (int i = 0; i < nums.length; i++) {
hashMap.put(nums[i], i);
}

int[] result = null;
for (int i = 0; i < nums.length; i++) {
int num = nums[i];
int res = target - num;
if (hashMap.containsKey(res) && hashMap.get(res) != i) {
result = new int[2];
result[0] = i;
result[1] = hashMap.get(res);
break;
}
}
return result;
}
}