給定一個有 n 個整數(shù)的數(shù)組 S,找出 S 中 3 個數(shù),使其和等于一個給定的數(shù),target。
返回這 3 個數(shù)的和,你可以假定每個輸入都有且只有一個結(jié)果。
例如,給定 S = {-1 2 1 -4},和 target = 1。
那么最接近 target 的和是 2。(-1 + 2 + 1 = 2)。
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).
也許我已經(jīng)開始體會到上一題中別人寫的方法的思想了。
在這個題目中,我們要做以下幾件事:
class Solution
{
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
int len = nums.size();
int result = INT_MAX, close = INT_MAX;
for (int current = 0; current < len - 2; current++) {
int front = current + 1, back = len - 1;
while (front < back) {
int sum = nums[current] + nums[front] + nums[back];
if (sum < target) {
if (target - sum < close) {
close = target - sum;
result = sum;
}
front++;
}
else if (sum > target) {
if (sum - target < close) {
close = sum - target;
result = sum;
}
back--;
}
else {
close = 0;
result = target;
do {
front++;
} while (front < back&&nums[front - 1] == nums[front]);
do {
back--;
} while (front < back&&nums[back + 1] == nums[back]);
}
}
while (current < len - 2 && nums[current + 1] == nums[current]) {
current++;
}
}
return result;
}
};
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