题目地址：https://leetcode.com/problems/two-sum/description/
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:
代码语言：javascript复制Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] nums[1] = 2 7 = 9,
return [0, 1].核心思想：
利用hashmap在遍历一次的情况下，先存储数组中数字和角标的对应关系和判断是否含有target-this。
代码语言：javascript复制 /**
* 整形数组，找出和为某数的两个数字
* @param nums 整型数组
* @param target 目标数字
* @return 满足和为target的两个整数
*/
public int[] twoSum2(int[] nums, int target) {
Map tmpMap = new HashMap<>(nums.length);
//遍历并记录到Map中
for (int i = 0; i < nums.length; i ) {
int complement = target - nums[i];
if (tmpMap.containsKey(complement)&&tmpMap.get(complement)!=i){
return new int[]{i,tmpMap.get(complement)};
}
tmpMap.put(nums[i],i);
}
throw new IllegalArgumentException("No two sum solution");
}
/**
* 整形数组，找出和为某数的两个数字
* @param nums 整型数组
* @param target 目标数字
* @return 满足和为target的两个整数
*/
public int[] twoSum(int[] nums, int target) {
Map tmpMap = new HashMap<>(nums.length);
//遍历并记录到Map中
for (int i = 0; i < nums.length; i ) {
tmpMap.put(nums[i],i);
}
//遍历一次直接匹配
for (int i = 0; i < nums.length; i ) {
if (tmpMap.containsKey(target - nums[i])&&tmpMap.get(target - nums[i])!=i){
return new int[]{i,tmpMap.get(target - nums[i])};
}
}
throw new IllegalArgumentException("No two sum solution");
}
