leetcode-1. Two Sum-簡單的演算法題,面試見到了嗎?
阿新 • • 發佈:2018-12-24
leetcode第一道題,很簡單大部分人應該能想到用map,而我只想到暴力,思維確實得到了鍛鍊。
Question: 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].
descript:給定一個數組,和目標值,求數組裡兩個數的和等於目標值的下標,數組裡數不重複。
菜鳥解法 (Brute Force)
public int[] twoSum(int[] nums, int target) {
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
if (nums[j] == target - nums[i]) {
return new int[] { i, j };
}
}
}
throw new IllegalArgumentException("No two sum solution");
}
*Time complexity : O(n2
)
Space complexity : O(1
)*
大鳥解法
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
map.put(nums[i], i);
}
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement) && map.get(complement) != i) {
return new int[] { i, map.get(complement) };
}
}
throw new IllegalArgumentException("No two sum solution");
*Time complexity : O(n)
Space complexity : O(n)*
飛鷹解法
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (map.containsKey(complement)) {
return new int[] { map.get(complement), i };
}
map.put(nums[i], i);
}
throw new IllegalArgumentException("No two sum solution");
}
Time complexity : O(n)
Space complexity : O(n)