add: 3Sum Closest

Author: hijiangtao

README.md

@@ -19,6 +19,7 @@ This is the solutions collection of my LeetCode submissions, most of them are pr
 |13|[Roman to Integer](https://leetcode.com/problems/roman-to-integer/) | [JavaScript](./src/roman-to-integer/res.js)|Easy|
 |14|[Longest Common Prefix](https://leetcode.com/problems/longest-common-prefix/) | [JavaScript](./src/longest-common-prefix/res.js)|Easy|
 |15|[3Sum](https://leetcode.com/problems/3sum/) | [JavaScript](./src/3sum/res.js) |Medium|
+|16|[3Sum Closest](https://leetcode.com/problems/3sum-closest/) | [JavaScript](./src/3sum-closest/res.js) |Medium|
 |17|[Letter Combinations of a Phone Number](https://leetcode.com/problems/letter-combinations-of-a-phone-number/) | [JavaScript](./src/letter-combinations-of-a-phone-number/res.js)|Medium|
 |19|[Remove Nth Node From End of List](https://leetcode.com/problems/remove-nth-node-from-end-of-list/) | [JavaScript](./src/remove-nth-node-from-end-of-list/res.js)|Medium|
 |22|[Generate Parentheses](https://leetcode.com/problems/generate-parentheses/) | [JavaScript](./src/generate-parentheses/res.js)|Medium|

src/3sum-closest/res.js

@@ -0,0 +1,49 @@
+/**
+ * res.js
+ * @authors Joe Jiang (hijiangtao@gmail.com)
+ * @date    2017-05-17 09:40:45
+ * 
+ * Solution: https://discuss.leetcode.com/topic/1978/a-n-2-solution-can-we-do-better/2
+ * 
+ * @param {number[]} nums
+ * @param {number} target
+ * @return {number}
+ */
+let threeSumClosest = function(nums, target) {
+    let len = nums.length;
+
+    if (len < 3) {
+        return nums.reduce(reduceArray, 0);
+    }
+    nums.sort(compareNumbers);
+
+    function compareNumbers(a, b) {
+        return a - b;
+    }
+
+    function reduceArray(a, b) {
+        return a + b;
+    }
+
+    let res = nums[0] + nums[1] + nums[2];
+    for (let i = 0; i < len - 2; i++) {
+        let j = i + 1,
+            k = len - 1;
+
+        while (j < k) {
+            let sum = nums[i] + nums[j] + nums[k];
+            if (Math.abs(res - target) > Math.abs(sum - target)) {
+                if (sum === target) return target;
+                res = sum;
+            }
+
+            if (sum > target) {
+                k--;
+            } else {
+                j++;
+            }
+        }
+    }
+
+    return res;
+};