精通算法系列-三值更小

原题:

// Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.// For example, given nums = [-2, 0, 1, 3], and target = 2.// Return 2. Because there are two triplets which sums are less than 2:// [-2, 0, 1]// [-2, 0, 3]// Follow up:    // Could you solve it in O(n2) runtime?

意思就是:选三个值满足0 <= i < j < k < n 且这三个值加起来没有给定的值大的数组,

答案如下:

public class Solution {    public int threeSumSmaller(int[] nums, int target) {        //initialize total count to zero        int count = 0;        //sort the array        Arrays.sort(nums);        //loop through entire array        for(int i = 0; i < nums.length - 2; i++) {            //set left to i + 1            int left = i + 1;            //set right to end of array            int right = nums.length - 1;            //while left index < right index            while(left < right) {                //if the 3 indices add to less than the target increment count                if(nums[i] + nums[left] + nums[right] < target) {                    //increment the count by the distance between left and right because the array is sorted                    count += right - left;                    //increment left pointer                    left++;                }                //if they sum to a value greater than target...                else {                    //decrement right pointer                    right--;                }            }        }        return count;    }}

核心思路:从左到左加1到最后开始循环，满足条件则加1