3Sum Closest (M)
題目
Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
Example:
Given array nums = [-1, 2, 1, -4], and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
題意
求一三元組，使其和最接近給定的值，
思路
與 0015. 3Sum 類似，只是修改了判定的標準，先排序，后利用two pointers并去除重復值即可，
代碼實作
Java
class Solution {
public int threeSumClosest(int[] nums, int target) {
int min = Integer.MAX_VALUE;
int ans = 0;
Arrays.sort(nums);
for (int i = 0; i < nums.length - 2; i++) {
// 去除重復的i
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int j = i + 1, k = nums.length - 1;
while (j < k) {
int sum = nums[i] + nums[j] + nums[k];
int diff = Math.abs(sum - target);
if (diff < min) {
min = diff;
ans = sum;
}
// 移動j、k的標準是使sum更加接近target
if (sum < target) {
j++;
while (j < k && nums[j] == nums[j - 1]) {
j++;
}
} else if (sum > target) {
k--;
while (j < k && nums[k] == nums[k + 1]) {
k--;
}
} else {
return sum; // 特殊情況，sum正好為target
}
}
}
return ans;
}
}
JavaScript
/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var threeSumClosest = function (nums, target) {
let closest = 0
let minDis = Number.MAX_SAFE_INTEGER
nums.sort((a, b) => a - b)
for (let i = 0; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue
}
let j = i + 1, k = nums.length - 1
while (j < k) {
if (j > i + 1 && nums[j] == nums[j - 1]) {
j++
} else if (k < nums.length - 1 && nums[k] == nums[k + 1]) {
k--
} else {
let sum = nums[i] + nums[j] + nums[k]
let dis = Math.abs(sum - target)
if (dis < minDis) {
minDis = dis
closest = sum
}
if (sum < target) j++
else if (sum > target) k--
else return sum
}
}
}
return closest
}
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