題目

Given an array?S?of?n?integers, are there elements?a,?b,?c, and?d?in?S?such that?a?+?b?+?c?+?d?= target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie,?a?≤?b?≤?c?≤?d)
• The solution set must not contain duplicate quadruplets.

For example, given array S = {1 0 -1 0 -2 2}, and target = 0.A solution set is:(-1, 0, 0, 1)(-2, -1, 1, 2)(-2, 0, 0, 2)

方法

將數(shù)組排序，從頭開(kāi)始。先選擇一個(gè)元素。剩余的元素轉(zhuǎn)換為求解三個(gè)數(shù)的和。

public List<List<Integer>> threeSum(int[] num, int start, int target) {List<List<Integer>> list = new ArrayList<List<Integer>>();int len = num.length;for (int i = start; i < len - 2; i++) {if (i == start || num[i] != num[i - 1]) {int tempTarget = target - num[i];int left = i + 1;int right = len - 1;while (left < right) {if (left < right) {int temp = num[left] + num[right];if (temp == tempTarget) {if (left == i + 1 || num[left] != num [left - 1]) {List<Integer> subList = new ArrayList<Integer>();subList.add(num[i]);subList.add(num[left]);subList.add(num[right]);list.add(subList);}left++;right--;} else if (temp > tempTarget) {right--;} else {left++;}}} }}return list;}public List<List<Integer>> fourSum(int[] num, int target) {List<List<Integer>> list = new ArrayList<List<Integer>>();Arrays.sort(num);int len = num.length;for (int i = 0; i < len - 3; i++) {if (i == 0 || num[i] != num[i - 1]) {int tempTarget = target - num[i];List<List<Integer>> tempList = threeSum(num, i + 1, tempTarget);for (int j = 0; j < tempList.size(); j++) {List<Integer> subList = tempList.get(j);List<Integer> newList = new ArrayList<Integer>(subList);newList.add(0, num[i]);list.add(newList);}}}return list;}

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