Given an array of integers, return indices of the two numbers such that they add up to a specific target.

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:

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Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Solution

From Ang’s:

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public int[] twoSum(int[] nums, int target) {
if(nums == null){
throw new IllegalArgumentException("Nums should not null");
}
int count = nums.length;
if(count == 0){
throw new IllegalArgumentException("Nums should not empty");
}
int i, j;
for(i = 0; i<count; i++){
for(j = i+1; j<count; j++){
if(nums[i] + nums[j] == target){
return new int[] { i, j };
}
}
}
throw new IllegalArgumentException("No two sum solution");
}

The better one (O(n)):

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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");
}