1. Two Sum
Easy
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, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
package leetcode;
import java.util.HashMap;
import java.util.Map;
public class TwoSum {
@org.junit.Test
public void test() {
int[] nums = { 2, 7, 11, 15 };
int target = 9;
Solution solution = new Solution();
int[] result = solution.twoSum1(nums, target);
System.out.println(result[0] + " " + result[1]);
result = solution.twoSum2(nums, target);
System.out.println(result[0] + " " + result[1]);
result = solution.twoSum3(nums, target);
System.out.println(result[0] + " " + result[1]);
}
}
class Solution {
public int[] twoSum1(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");
}
public int[] twoSum2(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");
}
public int[] twoSum3(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");
}
}
原文:https://www.cnblogs.com/denggelin/p/11517583.html