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.
给定一个整数数组，返回两个数字的索引，使它们相加到一个特定的目标。
您可以假设每个输入都只有一个解决方案，而您可能不会使用相同的元素两次。
Given nums = [2, 11, 7, 15], target = 9, Because nums[0] + nums[2] = 2 + 7 = 9, return [0, 2].思考：
1 可以先排序然后使用两个指针向中间遍历 时间复杂度O(nlongn)+O(n)
while(left < right){ int temp = nums[left] + nums[right]; if(temp == target){ break; }else if(temp < target){ left++; }else{ right--; } }
2 申请一个hashMap key记录数组元素 value数组元素下表 时间复杂度O(n)
for (int i = 0; i < numbers.length; i++) { if (map.containsKey(target - numbers[i])) { result[1] = i + 1; result[0] = map.get(target - numbers[i]); return result; } map.put(numbers[i], i + 1); }
转载于:https://www.cnblogs.com/WegYcx/p/7607157.html
