77 lines
2.0 KiB
C++
77 lines
2.0 KiB
C++
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#include <bits/stdc++.h>
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using namespace std;
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int n, m;
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const int N = 110;
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/*
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前提：q是一个有序递增的数组,当容器中的元素按照递增的顺序存储时，
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*/
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int q[N] = {1, 2, 3, 3, 3, 4, 5};
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int l, r;
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//二分算法之手动版本
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int manual_lower_bound(int l, int r, int x) {
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while (l < r) {
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int mid = (l + r) / 2;
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if (q[mid] >= x)
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r = mid;
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else
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l = mid + 1;
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}
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return l;
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}
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int manual_upper_bound(int l, int r, int x) {
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while (l < r) {
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int mid = (l + r) / 2;
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if (q[mid] > x)
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r = mid;
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else
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l = mid + 1;
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}
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return l;
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}
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int main() {
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/**********************************************************/
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//方法1：yxc 大法
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//从0~6找出>=3的第一个位置
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l = 0, r = 6;
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int x = 3;
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while (l < r) {
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int mid = (l + r) >> 1;
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if (q[mid] >= x)
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r = mid;
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else
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l = mid + 1;
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}
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printf("%d\n", l);
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//从0~6找出<=3的最后一个位置
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l = 0, r = 6;
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x = 3;
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while (l < r) {
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int mid = (l + r + 1) >> 1;
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if (q[mid] <= x)
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l = mid;
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else
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r = mid - 1;
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}
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printf("%d\n", l);
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/*
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方法2：STL lower_bound,upper_bound 大法
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lower_bound函数返回容器中第一个大于等于目标值的位置
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upper_bound函数返回容器中第一个大于目标值的位置
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若容器中的元素都比目标值小则返回最后一个元素的下一个位置
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*/
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printf("%lld\n", lower_bound(q, q + 7, x) - q);
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printf("%lld\n", upper_bound(q, q + 7, x) - q - 1);
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/**********************************************************/
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/*
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方法3：手写 lower_bound,upper_bound大法
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*/
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printf("%d\n", manual_lower_bound(0, 7, x));
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printf("%d\n", manual_upper_bound(0, 7, x) - 1);
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/**********************************************************/
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return 0;
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}
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