1.从一串数字中找到相加等于target的两个数。TWO SUM

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.

从一些系列数字中找出两个数相加等于target。只有一个结果，且计算过程中，每个数字元素只能被用两次？？

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,

return [0, 1].

——————————————————————————————（写法一）————————————————————————————————————

public int[] twoSum(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");
}

————————————————————————————————（写法二）——————————————————————————————————————

class Solution {
public:
vector<int> twoSum(vector<int> &nums, int target)
{
vector<int> result;
int N = nums.size();
for (int i = 0; i < N - 1; i++){
for (int j = i+1; j < N; j++){
if (nums[i] + nums[j] == target){
result.push_back(i);
result.push_back(j);
return result;
}
}
}
return result;
}
};

最后

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