- 1.兩數(shù)之和
- 11.盛水最多的容器
- 15.三數(shù)之和
- 16.最接近的三數(shù)之和
- 18.四數(shù)之和
雙指針?lè)ㄋ枷耄?/h3>
首先對(duì)數(shù)組進(jìn)行排序挽荡,使數(shù)組從小到大排列洪碳。確定數(shù)組中的一個(gè)數(shù)劝评,再將左指針l = i + 1因惭,右指針r = 數(shù)組長(zhǎng)度 - 1。分三種情況進(jìn)行判定潘拨,如果三數(shù)之和(l + r + i)相等于目標(biāo)值則采取相應(yīng)的操作吊输,如果三數(shù)之和大于目標(biāo)值,則r--(使三數(shù)之和減小铁追,因?yàn)閿?shù)組有序)季蚂,否則l++,增大三數(shù)之和琅束。(以三數(shù)之和為例)
三數(shù)之和(code)
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> lists = new ArrayList<>();
if (nums == null || nums.length < 3) return lists;
Arrays.sort(nums);
for (int i = 0; i < nums.length - 2; i++) {
if(nums[i] > 0) break; //如果nums[i] > 0 扭屁,則三數(shù)之和必大于0
if(i > 0 && nums[i] == nums[i - 1]) continue; //去重
int l = i + 1;
int r = nums.length - 1;
while (l < r) {
int sum = nums[i] + nums[l] + nums[r];
if (sum == 0) {
lists.add(Arrays.asList(nums[i], nums[l], nums[r]));
while(l < r && nums[l] == nums[l + 1]) l++;
while(l < r && nums[r] == nums[r - 1]) r--;
l++;
r--;
} else if (sum > 0) {
r--;
} else {
l++;
}
}
}
return lists;
}
}
四數(shù)之和(code)
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> lists = new ArrayList<>();
if (nums == null || nums.length < 4) return lists;
Arrays.sort(nums);
for (int i = 0; i < nums.length; i++) {
if(i > 0 && nums[i] == nums[i - 1]) continue;
for (int j = i + 1; j < nums.length; j++) {
if(j > i + 1 && nums[j] == nums[j - 1]) continue;
int l = j + 1;
int r = nums.length - 1;
while (l < r) {
int sum = nums[l] + nums[r] + nums[i] + nums[j];
if (sum == target) {
lists.add(Arrays.asList(nums[i], nums[j], nums[l], nums[r]));
while(l < r && nums[l] == nums[l + 1]) l++;
while(l < r && nums[r] == nums[r - 1]) r--;
l++;
r--;
} else if (sum > target) {
r--;
} else {
l++;
}
}
}
}
return lists;
}
}