1. Two Sum
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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
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
Thinking
返回数组 nums 中两个值的下标，这两个值的和等于所给的 target数组 nums 中的元素并不是有序的假定解是唯一的，并且每个元素不得重复使用需要特别注意元素是 target 的一半的情况还需要特别注意一个值重复出现的情况
Solution -1
public int[]
twoSum(
int[] nums,
int target) {
Map<Integer, Integer> map =
new HashMap<>();
for (
int i =
0; i < nums.length; i++) {
if ((target-nums[i]) == nums[i] && !map.containsKey(nums[i])) {
map.put(nums[i], -
1);
}
else {
map.put(nums[i], target-nums[i]);
}
}
int[] result =
new int[
2];
for (
int i =
0; i < nums.length; i++) {
if (map.containsKey(map.get(nums[i]))) {
result[
0] = i;
break;
}
}
for (
int i =
0; i < nums.length; i++) {
if (i != result[
0] && nums[i] == map.get(nums[result[
0]])) {
result[
1] = i;
break;
}
}
return result;
}
Solution -2
大神做法，借鉴学习
public int[]
twoSum(
int[] nums,
int target) {
Map<Integer, Integer> res =
new HashMap<>();
for (
int i =
0; i < nums.length; i++) {
int difference = target - nums[i];
if (res.containsKey(difference)) {
return new int[]{ res.get(difference), i };
}
res.put(nums[i], i);
}
return null;
}
