table of Contents
LeetCode16. The closest sum of three numbers
Hello, friends, I am your old friend Tsukahu. Today we will look at a medium-difficulty problem on LeetCode-the nearest sum of thirds problem. I hope that after reading this article, I can bring you some gains.
topic:
First, let's take a look at the description of the topic:
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analysis
For this problem, we first analyze from a simple scenario:
Just find a number
The simplest scenario is nothing more than finding the closest sum, then we only need to maintain one variable: best. Best records the current best value, and replaces the best when it is more suitable until the entire array is traversed.
Find the sum of two numbers
To be more complicated is to find the sum of the two closest numbers. The simplest idea is to reduce to the previous case, first subtract one value from the target value, and then look for the rest, but this way, the time complexity is not very friendly due to the nested for loop, and there is room for optimization.
Now the question is: Where is the room for optimization? How to save this part of space?
The problem with the above method is that when we are the first one that is already smaller than the target value, adding a negative value to it is even less likely to be our target. The next step is to expand, when we a+b When it is already less than the target value, it is completely unnecessary to find a value smaller than b to add to a.
- Ordered array plus double pointer:
- Then we sort the array first, and then we get an ordered array
- Then the left and right pointers point to the first and last respectively, add them to get sum, and then compare best and sum which is closer to the target value
- Then judge whether the sum is larger or smaller than the target value. If it is larger, the right pointer moves to the left, and if it is smaller, the left pointer moves to the right.
Find the sum of three numbers
Now that we have the experience of the first two times, then we will transform into finding the sum of two numbers, that is, find a fixed value first, and then double pointer.
Code
public int threeSumClosest(int[] nums, int target) {
// 先将数组排序以便和后续双指针运算
Arrays.sort(nums);
int len = nums.length;
// 赋值默认初始值，
int best = nums[0] + nums[1] + nums[2];
int sum = 0;
// 固定a找b和c
for (int a = 0; a < len; a++) {
// 为了防止重复查找 b=a+1;
int b = a + 1;
int c = len - 1;
while (b < c) {
sum = nums[a] + nums[b] + nums[c];
if (sum == target) {
return sum;
}
// 判断是否需要更新最佳值
if (Math.abs(sum - target) < Math.abs(best - target)) {
best = sum;
}
// 判断双指针哪一个需要移动
if (sum > target) {
c--;
} else {
b++;
}
}
}
return best;
}
At last
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