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leetcode/Arrays/README.md

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+Given an integer array nums of length n 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 1:
+
+Input: nums = [-1,2,1,-4], target = 1
+Output: 2
+Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
+Example 2:
+
+Input: nums = [0,0,0], target = 1
+Output: 0
+
+##### 단순히 3중 for문으로 풀었음...
+
+```java
+class Solution {
+ public int threeSumClosest(int[] nums, int target) { 
+
+ int num = nums[0] + nums[1] + nums[2];
+ int min = Math.abs(num-target);
+
+ for(int i=0; i<nums.length; i++) {
+ for(int j=i+1; j<nums.length; j++) {
+ int two = i+j;
+ for(int k=j+1; k<nums.length; k++) {
+ int temp = nums[i]+nums[j]+nums[k];
+ if(Math.abs(temp-target) < min) {
+ min = Math.abs(temp-target);
+ num = temp; 
+ }
+ }
+ }
+ }
+
+ return num;
+ }
+}
+```
+
+##### 가장 많은 추천을 받은 풀이, 하나만 고정해놓고 나머지 두개는 위 아래로 움직이네...
+```java
+public int threeSumClosest(int[] num, int target) {
+ int result = num[0] + num[1] + num[num.length - 1];
+ Arrays.sort(num);
+ for (int i = 0; i < num.length - 2; i++) {
+ int start = i + 1, end = num.length - 1;
+ while (start < end) {
+ int sum = num[i] + num[start] + num[end];
+ if (sum > target) {
+ end--;
+ } else {
+ start++;
+ }
+ if (Math.abs(sum - target) < Math.abs(result - target)) {
+ result = sum;
+ }
+ }
+ }
+ return result;
+ }
+```