3sum-closest.rs

#![allow(dead_code, unused, unused_variables)]

fnmain(){}

structSolution;

implSolution{
/// 暴力破解法
pubfnthree_sum_closest1(nums:Vec<i32>,target:i32) -> i32{
letmut sum:i32 = nums[0..3].iter().sum();

for i in0..nums.len(){
for j in i + 1..nums.len(){
for k in j + 1..nums.len(){
let s = nums[i] + nums[j] + nums[k];
if s == target {
return s;
}
 sum = if(sum - target).abs() > (s - target).abs(){
 s
}else{
 sum
};
}
}
}

 sum
}

/// 排序后，和是递增的
pubfnthree_sum_closest(nums:Vec<i32>,target:i32) -> i32{
letmut nums = nums;
 nums.sort();
letmut sum:i32 = nums[..3].iter().sum();

// 因为排序后递增，所以前三个数的和是最小的，当前三个数的和都大于target时，其他数的和肯定大于target
if sum > target {
return sum;
}

for i in0..nums.len(){
let(mut left,mut right) = (i + 1, nums.len() - 1);

while left < right {
let s = nums[i] + nums[left] + nums[right];
if s == target {
return s;
}elseif s > target {
 right -= 1;
}else{
 left += 1;
}

 sum = if(s - target).abs() > (sum - target).abs(){
 sum
}else{
 s
};
}
}

 sum
}
}