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问题描述
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2)
解决思路
1. 搜索(超时);
2. 排序 + 二分, 时间复杂度为O(n^2).
程序
public class ThreeSumClosest {
// dfs
private int diff = 0;
private int abs = Integer.MAX_VALUE;
public int threeSumClosest(int[] nums, int target) {
if (nums == null || nums.length < 3) {
return -1;
}
List<Integer> sol = new ArrayList<Integer>();
boolean[] used = new boolean[nums.length];
helper(nums, target, sol, used, 0);
return target - diff;
}
private void helper(int[] nums, int target, List<Integer> sol,
boolean[] used, int begin) {
if (sol.size() == 3) {
if (Math.abs(target) < abs) {
abs = Math.abs(target);
diff = target;
}
return;
}
for (int i = begin; i < nums.length; i++) {
if (used[i]) {
continue;
}
used[i] = true;
sol.add(nums[i]);
helper(nums, target - nums[i], sol, used, begin + 1);
used[i] = false;
sol.remove(sol.size() - 1);
}
}
// sort + bs
public int threeSumClosest(int[] nums, int target) {
if (nums == null || nums.length < 3) {
return -1;
}
Arrays.sort(nums);
int diff = 0;
int abs = Integer.MAX_VALUE;
for (int i = 0; i < nums.length - 2; i++) {
int begin = i + 1;
int end = nums.length - 1;
while (begin < end) {
int sum = nums[i] + nums[begin] + nums[end];
if (sum == target) {
return sum;
}
if (sum < target) {
++begin;
} else {
--end;
}
if (Math.abs(sum - target) < abs) {
abs = Math.abs(sum - target);
diff = sum - target;
}
}
}
return target + diff;
}
}
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原文地址:http://www.cnblogs.com/harrygogo/p/4675411.html