2017算法课.11(3Sum Closest)
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问题:
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).题意:
给定一个有n个整数的数组S,找出S中3个数,使其和等于一个给定的数,target。
返回这3个数的和,你可以假定每个输入都有且只有一个结果。
例如,给定S= {-1 2 1 -4},和target = 1。那么最接近target的和是2。(-1 + 2 + 1 = 2)。
在这个题目中,我们要做以下几件事:
用
sort 对输入的数组进行排序求出长度
len ,current 之所以要小于len−2 ,是因为后面需要留两个位置给front 和back 始终保证
front 小于back 计算索引为
current 、front 、back 的数的和,分别有比target 更小、更大、相等三种情况更小:如果距离小于
close ,那么close 便等于target−sum ,而结果就是sum 。更大的情况同理如果相等,那么要记得将
0 赋值给close ,result 就直接等于target 了随后为了避免计算重复的数字,用三个
do/while 循环递增或递减它们
代码:
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;
}
};
{
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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